Maja Andri´c, Ghulam Farid and Josip Peˇcari´ c

MONOGRAPHS IN INEQUALITIES 20

Analytical Inequalities for Fractional CalculusOperators and the Mittag-Leffler Function

Applications of integral operators containingan extended generalized Mittag-Leffler function in the kernel

Maja Andric, Ghulam Farid and Josip Pecaric

Analytical Inequalities forFractional Calculus Operatorsand the Mittag-Leffler Function

Applications of integral operators

containing an extended generalizedMittag-Leffler function in the kernel

Maja Andric

Faculty of Civil Engineering, Architecture and Geodesy

University of Split

Split, Croatia

Ghulam Farid

Department of Mathematics

COMSATS University Islamabad

Attock Campus, Pakistan

Josip Pecaric

RUDN University

Moscow, Russia.

Zagreb, 2021

MONOGRAPHS IN INEQUALITIES 20

Analytical Inequalities for Fractional CalculusOperators and the Mittag-Leffler Function

Maja Andric, Ghulam Farid and Josip Pecaric

Consulting Editors

Kristina Krulic Himmerlich

Faculty of Textile TechnologyUniversity of ZagrebZagreb, Croatia

Tibor Poganj

Faculty of Maritime StudiesUniversity of RijekaRijeka, Croatia

1st edition

Copyright c© by Element, Zagreb, 2021.

Printed in Croatia.All rights reserved.

A CIP catalogue record for this book is available from the National andUniversity Library in Zagreb under 001102208.

ISBN 978-953-197-813-2

No part of this publication may be reproduced or distributed in any formor by any means, or stored in a data base or retrieval system, withoutthe prior written permission of the publisher.

Preface

Fractional calculus is a theory of differential and integral operators of non-integer order.In recent years, considerable interest in this theory has been stimulated due to its manyapplications in almost all applied science, especially in numerical analysis and variousfields of physics and engineering. Fractional calculus enabled the adoption of a theoreticalmodel based on experimental data.

Inequalities which involve integrals of functions and their derivatives, whose study hasa history of about a century, are of great importance in mathematics, with far-reachingapplications in the theory of differential equations, approximations and probability, amongothers. They occupy a central place in mathematical analysis and its applications.

Fractional differentiation inequalities have applications to fractional differential equa-tions; the most important ones are in establishing uniqueness of the solution of initialproblems and giving upper bounds to their solutions. These applications have motivatedmany researchers in the field of integral inequalities to investigate certain extensions andgeneralizations using different fractional differential and integral operators. There are sev-eral well-known forms of fractional operators: Riemann-Liouville, Weyl, Erdelyi-Kober,Hadamard, Katugampola are just a few. All these forms of fractional operators in a specialcase are reduced to the Riemann-Liouville fractional integrals Ja+ f of order defined asin [92, 135] for f ∈ L1[a,b].

As a solution of fractional order differential or integral equations, the Mittag-Lefflerfunction with its generalizations appears. It is a function of one parameter defined by thepower series using the gamma function

E(z) =

n=0

zn

(n+1), (z, ∈ C ,() > 0)

and it is a natural extension of the exponential, hyperbolic and trigonometric functions.Extensions and generalizations of the Mittag-Leffler function enabled researchers to ob-tain fractional integral inequalities of different types, for example inequalities of the Opial,Polya-Szego, Chebyshev, Hermite-Hadamard and Fejer types, etc. Consequently, new re-sults are produced for more generalized fractional integral operator containing the Mittag-Leffler function in their kernels.

v

The book is divided in nine chapters. The first chapter presents notation, terms andsome important results for continuous and absolutely continuous functions. Definitionsand properties of different types of convex functions that will be used in the book aregiven, as well as an overview of fractional calculus.

In Chapter 2 we define the extended and generalized Mittag-Leffler function and giveits properties. For different parameter choices, corresponding known generalizations ofthe Mittag-Leffler function can be deduced: the Wiman generalization, also known asMittag-Leffler function of two parameters, Prabhakar’s function, Shukla-Prajapati’s func-tion, Salim-Faraj’s function or recent extension defined by Rahman et al. We also presentthe corresponding generalized fractional integral operators containing our extended gener-alized Mittag-Leffler function in the kernel.

Motivated by Opial type inequalities, in Chapter 3 we use the extended generalizedMittag-Leffler function with the corresponding fractional integral operator (in real domain)to obtain fractional generalizations of Opial type inequalities due to Mitrinovic and Pecaric.For such inequalities we construct functionals and give their mean value theorems.

In Chapter 4, we present improved and generalized Polya-Szego and Chebyshev typesfractional integral inequalities that are related to the Mittag-Leffler function. We also useKaramata’s estimations of the Chebyshev quotient to obtain even better upper and lowerestimations.

Chapter 5 is dedicated to Minkowsky type inequalities. We present fractional gener-alizations of integral inequality and its reverse versions for generalized fractional integraloperators containing our extended generalized Mittag-Leffler function.

Certain classical integral inequalities are presented in Chapter 6. We apply the resultsfrom Chapter 2 to obtain improvements, extensions and generalizations of known inequal-ities.

Famous Hadamard and Fejer-Hadamard inequalities are the main objects of Chap-ter 7, where we give Hadamard and Fejer- Hadamard types fractional inequalities for con-vex, relative convex, m-convex, (h−m)-convex, harmonically convex and harmonically(,h−m)-convex functions, which include our extended generalized Mittag-Leffler func-tion.

In Chapter 8, fractional integral inequalities are given that provide the bounds of var-ious kinds of fractional integral operators containing our extended generalized Mittag-Leffler function. We also give estimations of these inequalities for different kinds of con-vex functions. Results are applied on a certain function to establish recurrence relationsamong Mittag-Leffler functions.

The last chapter begins with the bounds of unified integral operators given for convexfunctions, continuing with results for the exponentially (s,m)-convex, strongly(s,m)-convex and (,m)-convex functions, which contain the Mittag-Leffler function.

Authors

vi

Contents

1 Preliminaries 1

1.1 Continuous and Absolutely Continuous Functions . . . . . . . . . . . . . 1

1.2 The Gamma and Beta Functions . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Convex Functions and Classes of Convexity . . . . . . . . . . . . . . . . 5

1.4 Fractional Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.4.1 The Riemann-Liouville Fractional Integrals and Derivatives . . . . 13

2 An Extended Generalized Mittag-Leffler Function 17

2.1 A Further Extension of the Mittag-Leffler Function . . . . . . . . . . . . 18

2.2 Fractional Integral Operators Associated withthe Mittag-Leffler Function . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.3 Unified Integral Operators . . . . . . . . . . . . . . . . . . . . . . . . . 26

3 Opial Type Fractional Integral Inequalities Associatedwith the Mittag-Leffler Function 29

3.1 Opial Type Fractional Integral Inequalities and an ExtendedGeneralized Mittag-Leffler Function . . . . . . . . . . . . . . . . . . . . 31

3.2 Properties of Associated Fractional Linear Functional . . . . . . . . . . . 35

4 Polya-Szego and Chebyshev Types Fractional Integral InequalitiesAssociated with the Mittag-Leffler Function 37

4.1 Polya-Szego Type Fractional Integral Inequalitiesand an Extended Generalized Mittag-Leffler Function . . . . . . . . . . . 38

4.2 Chebyshev Type Fractional Integral Inequalitiesand an Extended Generalized Mittag-Leffler Function . . . . . . . . . . . 42

vii

5 Minkowski Type Fractional Integral Inequalities Associatedwith the Mittag-Leffler Function 49

5.1 Reverse Minkowski Type Inequalities Involving an ExtendedGeneralized Mittag-Leffler Function . . . . . . . . . . . . . . . . . . . . 50

5.2 Related Minkowski Type Inequalities . . . . . . . . . . . . . . . . . . . 52

5.3 Further Generalizations of Minkowski Type Inequalities . . . . . . . . . 56

6 Classical Integral Inequalities and the Mittag-Leffler Function 63

6.1 Generalizations of Classical Integral Inequalities Involvingan Extended Generalized Mittag-Leffler Function . . . . . . . . . . . . . 64

6.2 Extensions of Classical Integral Inequalities Involvingan Extended Generalized Mittag-Leffler Function . . . . . . . . . . . . . 69

6.3 Further Generalizations of Some Classical Integral Inequalities . . . . . . 74

7 Hadamard and Fejer-Hadamard Types Fractional IntegralInequalities Associated with the Mittag-Leffler Function 81

7.1 Hadamard and Fejer-Hadamard Inequalities for Convex Functions . . . . 81

7.2 Hadamard and Fejer-Hadamard Inequalities for RelativeConvex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

7.3 Hadamard and Fejer-Hadamard Inequalities for m-convex Functions . . . 93

7.4 Hadamard and Fejer-Hadamard Inequalitiesfor (h−m)-convex Functions . . . . . . . . . . . . . . . . . . . . . . . . 100

7.5 Hadamard and Fejer-Hadamard inequalitiesfor Harmonically Convex Functions . . . . . . . . . . . . . . . . . . . . 104

7.6 Hadamard and Fejer-Hadamard Inequalitiesfor Harmonically (,h−m)-convex Functions . . . . . . . . . . . . . . . 122

7.6.1 Results for Harmonically (h−m)-convex Functions . . . . . . . 129

7.6.2 Results for Harmonically (,m)-convex Functions . . . . . . . . 131

8 Error Bounds of Hadamard and Fejer-Hadamard Inequalities and Boundsof Fractional Integral Operators Associated with Mittag-Leffler Function 133

8.1 Error Bounds Associated with Fractional Integral Inequalitiesfor Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

8.2 Error Bounds Associated with Fractional Integral Inequalitiesfor m-convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

8.3 Bounds of Fractional Integral Operators for (h−m)-convex Functions . . 154

8.4 Inequalities for the Extended Generalized Mittag-Leffler Function . . . . 160

viii

8.5 Error Bounds of Fractional Integral Operatorsfor Quasi-convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . 163

8.5.1 Recurrence Inequalities for Mittag-Leffler Functions . . . . . . . 164

8.5.2 Error Bounds of Hadamard and Fejer-Hadamard Inequalities . . 167

8.6 Bounds of Fractional Integral Operators for s-convex Functions . . . . . . 172

8.7 Bounds of Fractional Integral Inequalities for (s,m)-convex Functions . . 178

8.8 Bounds of Fractional Integral Operatorsfor Exponentially s-convex Functions . . . . . . . . . . . . . . . . . . . 185

8.9 Bounds of Fractional Integral Operatorsfor Strongly (s,m)-convex Functions . . . . . . . . . . . . . . . . . . . . 192

8.10 Bounds of Fractional Integral Operatorsfor Exponentially (s,m)-convex Functions . . . . . . . . . . . . . . . . . 204

8.11 Bounds of Fractional Integral Operatorsfor Exponentially m-convex Functions . . . . . . . . . . . . . . . . . . . 210

9 Bounds of Unified Integral Operators Containing Mittag-Leffler Function 225

9.1 Bounds of Unified Integral Operators of Convex Functions . . . . . . . . 225

9.2 Bounds of Unified Integral Operatorsfor Exponentially (s,m)-Convex Functions . . . . . . . . . . . . . . . . . 234

9.3 Bounds of Unified Integral Operatorsfor Strongly (s,m)-Convex Functions . . . . . . . . . . . . . . . . . . . 245

9.4 Bounds of Unified Integral Operators for (,m)-Convex Functions . . . . 253

Bibliography 261

Index 271

ix

Chapter1

Preliminaries

1.1 Continuous and Absolutely Continuous Functions

We start with definitions and properties of integrable functions, continuous functions, ab-solutely continuous functions, and give required notation, terms and overview of someimportant results (more details could be found in monographs [106, 133]).

Lp spaces

Let [a,b] be a finite interval in R, where−≤ a < b≤. We denote by Lp[a,b], 1≤ p <,the space of all Lebesgue measurable functions f for which

∫ ba | f (t)|p dt < , with the

norm

|| f ||p =(∫ b

a| f (t)|p dt

) 1p

endowed, and by L[a,b] the set/space of all functions measurable and essentially boundedon [a,b], equipped with the norm

|| f || = esssup{| f (x)| : x ∈ [a,b]} .

1

2 1 PRELIMINARIES

Spaces of continuous and absolutely continuous functions

We denote by Cn[a,b], n ∈ N0, the space of functions which are n times continuouslydifferentiable on [a,b], that is

Cn[a,b] ={

f : [a,b]→ R : f (k) ∈C[a,b] ,k = 0,1, . . . ,n}

.

In particular, C0[a,b] =C[a,b] is the space of continuous functions on [a,b] with the norm

|| f ||Cn =n

k=0

|| f (k)||C =n

k=0

maxx∈[a,b]

| f (k)(x)| ,

and for C[a,b]|| f ||C = max

x∈[a,b]| f (x)| .

Lemma 1.1 The space Cn[a,b] consists of those and only those functions f which arerepresented in the form

f (x) =1

(n−1)!

∫ x

a(x− t)n−1(t)dt +

n−1

k=0

ck(x−a)k , (1.1)

where ∈C[a,b] and ck are arbitrary constants (k = 0,1, . . . ,n−1).Moreover,

(t) = f (n)(t) , ck =f (k)(a)

k!(k = 0,1, . . . ,n−1) . (1.2)

The space of absolutely continuous functions on a finite interval [a,b] is denoted byAC[a,b]. It is known that AC[a,b] coincides with the space of primitives of Lebesgueintegrable functions L1[a,b] (see Kolmogorov and Fomin [93, Chapter 33.2]):

f ∈ AC[a,b] ⇔ f (x) = f (a)+∫ x

a(t)dt , ∈ L1[a,b] ,

and therefore an absolutely continuous function f has an integrable derivative f ′(x) = (x)almost everywhere on [a,b]. We denote by ACn[a,b], n ∈ N, the space

ACn[a,b] ={

f ∈Cn−1[a,b] : f (n−1) ∈ AC[a,b]}

.

In particular, AC1[a,b] = AC[a,b].

Lemma 1.2 The space ACn[a,b] consists of those and only those functions which can berepresented in the form (1.1), where ∈ L1[a,b] and ck are arbitrary constants(k = 0,1, . . . ,n−1).Moreover, (1.2) holds.

1.2 THE GAMMA AND BETA FUNCTIONS 3

Theorem 1.1 (FUBINI’S THEOREM) Let (X ,M ,) and (Y,N ,) be -finite measurespaces and f ×-measurable function on X ×Y. If f ≥ 0, then next integrals are equal∫

X×Yf (x,y)d(×)(x,y),

∫X

(∫Y

f (x,y)d(y))

d(x)

and ∫Y

(∫X

f (x,y)d(x))

d(y).

If f is a complex function, then above equalities hold with additional requirement∫X×Y

| f (x,y)|d(×)(x,y) < .

Fubini’s theorem and its consequences below have numerous applications involving mul-tiple integrals: ∫ b

adx∫ d

cf (x,y)dy =

∫ d

cdy∫ b

af (x,y)dx;

∫ b

adx∫ x

af (x,y)dy =

∫ b

ady∫ b

yf (x,y)dx. (1.3)

1.2 The Gamma and Beta Functions

The gamma function is the function of complex variable defined by Euler’s integral ofsecond kind

(z) =∫

0tz−1 e−t dt , (z) > 0 . (1.4)

This integral is convergent for each z ∈ C such that (z) > 0. It has next property

(z+1) = z(z) , (z) > 0 ,

from which follows(n+1) = n! , n ∈ N0 .

For domain (z) ≤ 0 we have

(z) =(z+n)

(z)n, (z) > −n; n ∈ N; z �∈ Z

−0 = {0,−1,−2, . . .} , (1.5)

where (z)n is the Pochhammer’s symbol defined for z ∈ C and n ∈ N0 by

(z)0 = 1;

4 1 PRELIMINARIES

(z)n =(z+n)(z)

= z(z+1) · · ·(z+n−1), n ∈ N . (1.6)

The generalized Pochhammer’s symbol is defined for z, ∈ C by

(z) =(z+)(z)

. (1.7)

The gamma function is analytic in complex plane except in 0,−1,−2, . . . which are simplepoles. Another interesting equality holds:

(z)m+n = (z+m)n (z)m , n,m ∈ N .

The beta function is the function of two complex variables defined by Euler’s integralof the first kind

B(z,w) =∫ 1

0tz−1 (1− t)w−1dt , (z),(w) > 0 . (1.8)

It is related to the gamma function with

B(z,w) =(z)(w)(z+w)

, z,w �∈ Z−0 = {0,−1,−2, . . .} ,

which gives

B(z+1,w) =z

z+wB(z,w) .

Further, we have an extension of the beta function (for more details see [31])

Bp(x,y) =∫ 1

0tx−1(1− t)y−1e

− pt(1−t) dt, (x),(y),(p) > 0. (1.9)

Here we emphasize two equalities for the extended beta function:

Bp(x,y+1)+Bp(x+1,y) = Bp(x,y) ,∫

0Bp(x,y)dp = B(x+1,y+1), (x),(y) > −1.

Following examples of integrals will be often used in proofs and calculations in this book.

Example 1.1 Let , > 0 and x ∈ [a,b]. Then by substitution t = x− s(x−a) we have

∫ x

a(x− t)−1(t−a)−1dt =

∫ 1

0(x−a)+−1 s−1(1− s)−1ds

= B(, )(x−a)+−1 .

Analogously, by substitution t = x+ s(b− x), it follows

∫ b

x(t − x)−1(b− t)−1dt = B(, )(b− x)+−1 .

1.3 CONVEX FUNCTIONS AND CLASSES OF CONVEXITY 5

Example 1.2 Let , > 0, f ∈ L1[a,b] and x ∈ [a,b]. Then interchanging the order ofintegration and evaluating the inner integral we obtain∫ x

a(x− t)−1

∫ t

a(t− s)−1 f (s)dsdt =

∫ x

s=af (s)

∫ x

t=s(x− t)−1(t− s)−1 dt ds

= B(, )∫ x

a(x− s)+−1 f (s)ds .

Analogously,∫ b

x(t− x)−1

∫ b

t(s− t)−1 f (s)dsdt = B(, )

∫ b

x(s− x)+−1 f (s)ds .

1.3 Convex Functions and Classes of Convexity

Definitions and properties of convex functions, with more details, could be found in mono-graphs [107, 110, 121].

Let I be an interval in R.

Definition 1.1 A function f : I → R is called convex if

f ((1− )x+y)≤ (1− ) f (x)+ f (y) (1.10)

for all points x and y in I and all ∈ [0,1]. It is called strictly convex if the inequality(1.10) holds strictly whenever x and y are distinct points and ∈ (0,1). If − f is convex(respectively, strictly convex) then we say that f is concave (respectively, strictly concave).If f is both convex and concave, then f is said to be affine.

Lemma 1.3 (THE DISCRETE CASE OF JENSEN’S INEQUALITY) A real-valued functionf defined on an interval I is convex if and only if for all x1, . . . ,xn in I and all scalars1, . . . ,n in [0,1] with n

k=1k = 1 we have

f

(n

k=1

kxk

)≤

n

k=1

k f (xk) . (1.11)

The above inequality is strict if f is strictly convex, all the points xk are distinct and allscalars k are positive.

Theorem 1.2 (JENSEN) Let f : I → R be a continuous function. Then f is convex if andonly if f is midpoint convex, that is,

f

(x+ y

2

)≤ f (x)+ f (y)

2(1.12)

for all x,y ∈ I.

6 1 PRELIMINARIES

Corollary 1.1 Let f : I → R be a continuous function. Then f is convex if and only if

f (x+h)+ f (x−h)−2 f (x)≥ 0 (1.13)

for all x ∈ I and all h > 0 such that both x+h and x−h are in I.

Proposition 1.1 (THE OPERATIONS WITH CONVEX FUNCTIONS)

(i) The addition of two convex functions (defined on the same interval) is a convex func-tion; if one of them is strictly convex, then the sum is also strictly convex.

(ii) The multiplication of a (strictly) convex function with a positive scalar is also a(strictly) convex function.

(iii) The restriction of every (strictly) convex function to a subinterval of its domain isalso a (strictly) convex function.

(iv) If f : I → R is a convex (respectively a strictly convex) function and g : R → R isa nondecreasing (respectively an increasing) convex function, then g ◦ f is convex(respectively strictly convex)

(v) Suppose that f is a bijection between two intervals I and J. If f is increasing, thenf is (strictly) convex if and only if f−1 is (strictly) concave. If f is a decreasingbijection, then f and f−1 are of the same type of convexity.

Definition 1.2 If g is strictly monotonic, then f is said to be (strictly) convex with respectto g if f ◦ g−1 is (strictly) convex.

Proposition 1.2 If x1,x2,x3 ∈ I are such that x1 < x2 < x3, then the function f : I → R

is convex if and only if the inequality

(x3− x2) f (x1)+ (x1− x3) f (x2)+ (x2− x1) f (x3) ≥ 0

holds.

Proposition 1.3 If f is a convex function on an interval I and if x1 ≤ y1, x2 ≤ y2, x1 �= x2,y1 �= y2, then the following inequality is valid

f (x2)− f (x1)x2− x1

≤ f (y2)− f (y1)y2− y1

.

If the function f is concave, then the inequality reverses.

The following theorems concern derivatives of convex functions.

Theorem 1.3 Let f : I → R be convex. Then

(i) f is Lipschitz on any closed interval in I (recall, a function f such that | f (x)−f (y)| ≤ C|x− y| for all x and y, where C is a constant independent of x and y, iscalled a Lipshitz function);


(ii) f ′+ and f ′− exist and are increasing in I, and f ′− ≤ f ′+ (if f is strictly convex, thenthese derivatives are strictly increasing);

(iii) f ′ exists, except possibly on a countable set, and on the complement of which it iscontinuous.

Proposition 1.4 Suppose that f : I → R is a twice differentiable function. Then

(i) f is convex if and only if f ′′ ≥ 0;

(ii) f is strictly convex if and only if f ′′ ≥ 0 and the set of points where f ′′ vanishes doesnot include intervals of positive length.

Next we need divided differences, commonly used when dealing with functions thathave different degree of smoothness.

Definition 1.3 Let f : I →R, n∈N0 and let x0,x1, . . . ,xn ∈ I be mutually different points.The n-th order divided difference of a function at x0, . . . ,xn is defined recursively by

[xi; f ] = f (xi) , i = 0,1, . . . ,n ,

[x0,x1; f ] =[x0; f ]− [x1; f ]

x0− x1=

f (x0)− f (x1)x0 − x1

,

[x0,x1,x2; f ] =[x0,x1; f ]− [x1,x2; f ]

x0− x2, (1.14)

...

[x0, . . . ,xn; f ] =[x0, . . . ,xn−1; f ]− [x1, . . . ,xn; f ]

x0− xn.

Remark 1.1 The value [x0,x1,x2; f ] is independent of the order of the points x0, x1 andx2. This definition may be extended to include the case in which some or all the pointscoincide. Namely, taking the limit x1 → x0 in (1.14), we get

limx1→x0

[x0,x1,x2; f ] = [x0,x0,x2; f ] =f (x0)− f (x2)− f ′(x0)(x0 − x2)

(x0 − x2)2 , x2 �= x0

provided that f ′ exists, and furthermore, taking the limits xi → x0, i = 1,2 in (1.14), weget

limx2→x0

limx1→x0

[x0,x1,x2; f ] = [x0,x0,x0; f ] =f ′′(x0)

2

provided that f ′′ exists.

Definition 1.4 A function f : I → R is said to be n-convex (n ∈ N0) if for all choices ofn+1 distinct points x0, . . . ,xn ∈ I, the n-th order divided difference of f satisfies

[x0, . . . ,xn; f ] ≥ 0 . (1.15)

8 1 PRELIMINARIES

Thus the 1-convex functions are the nondecreasing functions, while the 2-convex functionsare precisely the classical convex functions.

Definition 1.5 A function f : I → (0,) is called log-convex if

f ((1− )x+y)≤ f (x)1− f (y) (1.16)

for all points x and y in I and all ∈ [0,1].

If a function f : I → R is log-convex, then it is also convex, which is a consequence of theweighted AG-inequality.

We continue with definitions and properties of other types of convex functions that willbe used in the book.

Definition 1.6 [112] Let Tg be a set of real numbers. This set Tg is said to be relativeconvex with respect to an arbitrary function g : R → R if

(1− t)x+ tg(y)∈ Tg

where x,y ∈ R such that x,g(y) ∈ Tg, t ∈ [0,1].

Definition 1.7 [112] A function f : Tg → R is said to be relative convex, if there existsan arbitrary function g : R → R such that

f ((1− t)x+ tg(y))≤ (1− t) f (x)+ t f (g(y)),

holds, where x,y ∈ R such that x,g(y) ∈ Tg, t ∈ [0,1].

Note that if g is identity function, then convex set and convex function are reproduced fromrelative convex set and relative convex function.

Definition 1.8 [73] Let I be an interval of real numbers. Then a function f : I → R issaid to be quasi-convex function, if for all a,b ∈ I and 0 ≤ t ≤ 1 the following inequalityholds

f (ta+(1− t)b)≤ max{ f (a), f (b)}. (1.17)

Example 1.3 [80] The function f : [−2,2]→ R, given by

f (x) ={

1 x ∈ [−2,−1]x2 x ∈ (−1,2]

is not a convex function on [−2,2], but it is quasi-convex function on [−2,2].

Definition 1.9 Let I be an interval of non-zero real numbers. A function f : I → R is saidto be harmonically convex function, if

f

(ab

ta+(1− t)b

)≤ t f (b)+ (1− t) f (a) (1.18)

holds for all a,b ∈ I and t ∈ [0,1]. If inequality in (1.18) is reversed, then f is said to beharmonically concave function.


Example 1.4 [76] Let f : (0,) → R be defined by f (t) = t and g : (−,0) → R de-fined by g(t) = t. Then f is harmonically convex function and g is harmonically concavefunction.

Following results are obvious from above example.

(i) If I ⊂ (0,) and f is non-decreasing convex function, then f is harmonically convex.

(ii) If I ⊂ (0,) and f is non-increasing harmonically convex function, then f is convex.

(iii) If I ⊂ (−,0) and f is non-decreasing harmonically convex function, then f is con-vex.

(iv) If I ⊂ (−,0) and f is non-increasing convex function, then f is harmonically con-vex.

Definition 1.10 [95] A function h : [a,b] ⊂ R \ {0} → R is said to be harmonicallysymmetric about a+b

2ab if for all x ∈ [a,b]

h

(1x

)= h

(1

1a + 1

b − x

).

We give another new notion of harmonically (h−m)-convex function by setting = 1as follows:

Definition 1.11 Let h : [0,1]⊆J→R be a nonnegative function. A function f :J⊆R+→R

is said to be harmonically (h−m)-convex if

f

(mxy

mty+(1− t)x

)≤ h(t) f (x)+mh(1− t) f (y)

holds for all x,y ∈ J, t ∈ [0,1] and m ∈ (0,1].

Next is a harmonically (,h−m)-convex function:

Definition 1.12 Let h : [0,1]⊆J→R be a nonnegative function. A function f :J⊆R+→R

is said to be harmonically (,h−m)-convex if

f

(mxy

mty+(1− t)x

)≤ h(t) f (x)+mh(1− t) f (y),

holds for all x,y ∈ J, t, ∈ [0,1] and m ∈ (0,1].

This unifies the definitions of harmonically ( ,m)-convexity and harmonically h-convexityof functions. For different specific choices of ,h,m, almost all kinds of well-knownharmonically convex functions can be obtained:

10 1 PRELIMINARIES

Remark 1.2 (i) If h(t) = t, then harmonically ( ,m)-convex function can be obtained[74].

(ii) If = 1 and h(t) = ts, then harmonically (s,m)-convex function can be obtained[27].

(iii) If = 1 and h(t) = t, then harmonically m-convex function can be obtained [27].

(iv) If = h(t) = m = 1, then harmonically P-function can be obtained [113].

(v) If = 1, h(t) = ts and m = 1, then harmonically s-convex function can be obtained[113].

(vi) If = 1, h(t) = 1t and m = 1, then harmonically Godunova-Levin function can be

obtained [113].

(vii) If = 1, h(t) = 1ts and m = 1, then harmonically s-Godunova-Levin function can be

obtained [113].

(viii) If we set m = 1 and = 1, then harmonically h-convex function can be achieved[113].

(ix) By putting = 1, h(t) = t and m = 1, then harmonically-convex function can beobtained [76].

Definition 1.13 [147] A function f : [0,b] → R, b > 0 is said to be m-convex function iffor all x,y ∈ [0,b] and t ∈ [0,1]

f (tx+m(1− t)y)≤ t f (x)+m(1− t) f (y)

holds where m ∈ [0,1].

Example 1.5 [105] A function f : [0,] → R given by

f (x) =112

(x3 −5x2 +9x−5x)

is 1617 -convex function. If m ∈ ( 16

17 ,1], then f is not m-convex.

For m = 1 the m-convex function reduces to convex function and for m = 0 it gives star-shaped function. If set of m-convex functions on [0,b] for which f (0) < 0 is denoted byKm(b), then we have

K1(b) ⊂ Km(b) ⊂ K0(b)

whenever m ∈ (0,1). In the class K1(b) there are convex functions f : [0,b]→ R for whichf (0) ≤ 0 (see, [40]).

Definition 1.14 [130] A function f : [a,b] → R is said to be exponentially m-convex if

e f (zx+m(1−z)y) ≤ ze f (x) +m(1− z)e f (y). (1.19)

for all x,y ∈ [a,b] and z ∈ [0,1] where m ∈ (0,1].


Definition 1.15 Let J ⊆R be an interval containing (0,1) and let h : J →R be a nonneg-ative function. We say that f : [0,b]→ R is a (h−m)−convex function, if f is nonnegativeand for all x,y ∈ [0,b], m ∈ [0,1] and ∈ (0,1), one has

f (x+m(1−)y)≤ h() f (x)+mh(1−) f (y).

For suitable choices of h and m, class of (h−m)−convex functions is reduces to differentknown classes of convex and related functions defined on [0,b] given in the followingremark:

Remark 1.3

(i) If m = 1, then we get h−convex function.

(ii) If h() = , then we get m−convex function.

(iii) If h() = and m = 1, then we get convex function.

(iv) If h() = 1 and m = 1, then we get p−function.

(v) If h() = s and m = 1, then we get s−convex function of second sense.

(vi) If h() = 1 and m = 1, then we get Godunova-Levin function.

(vii) If h() = 1s and m = 1, then we get s−Godunova-Levin function of second kind.

Definition 1.16 [103] A function f : [0,b]→ R, b > 0 is said to be (,m)-convex, where(,m) ∈ [0,1]2 if

f (tx+m(1− t)y)≤ t f (x)+m(1− t) f (y) (1.20)

holds for all x,y ∈ [0,b] and t ∈ [0,1].

Remark 1.4 (i) If we set = 1, then (1.20) gives the definition of m-convex function.

(ii) If we put (,m) = (1,1), then (1.20) gives the definition of convex function.

(iii) If we put (,m) = (1,0), then (1.20) gives the definition of starshaped function.

Definition 1.17 [71] Let s ∈ [0,1]. A function f : [0,) → R is said to be s-convexfunction in the second sense if

f (ta+(1− t)b)≤ ts f (a)+ (1− t)s f (b),

holds for all a,b ∈ [0,) and t ∈ [0,1].

Definition 1.18 [102] Let s ∈ (0,1] and I ⊆ [0,) be an interval. A function f : I → R issaid to be exponentially s-convex in the second sense if

f (ta+(1− t)b)≤ tsf (a)ea +(1− t)s f (b)

eb , (1.21)

holds for all a,b ∈ I, t ∈ [0,1] and ∈ R. If the inequality in (1.21) is reversed, then f iscalled exponentially s-concave.

12 1 PRELIMINARIES

Definition 1.19 [7] A function f : [0,b] → R, b > 0 is said to be (s,m)-convex functionwhere (s,m) ∈ [0,1]2, if

f (ta+m(1− t)b)≤ ts f (a)+m(1− ts) f (b)

for all x,y ∈ [0,b] and t ∈ [0,1].

Definition 1.20 [125] Let s ∈ [0,1] and I ⊆ [0,) be an interval. A function f : I → R issaid to be exponentially (s,m)-convex in second sense, if

f (tx+m(1− t)y)≤ tsf (x)ex +m(1− t)s f (y)

ey (1.22)

for all m ∈ (0,1] and ∈ R.

Remark 1.5

(i) For m = 1, (1.22) produces the definition of exponentially s-convex function.

(ii) For = 0, (1.22) produces the definition of (s,m)-convex function.

(iii) For = 0 and m = 1, (1.22) produces the definition of s-convex function.

(iv) For = 0 and m = 1, (1.22) produces the definition of convex function.

(v) For = 0 and s = 1, (1.22) produces the definition of m-convex function.

Definition 1.21 [75] Let I ⊂ (0,) be a real interval and p ∈ R \ {0}. Then a functionf : I → R is said to be p-convex, if

f([tap +(1− t)bp]

1p

)≤ t f (a)+ (1− t) f (b)

for all a,b ∈ I and t ∈ [0,1]. Note that for p = 1 and p = −1, p-convexity reduces toordinary convexity and harmonically convexity, respectively.

Definition 1.22 [94] Let p ∈ R\ {0}. Then a function f : [a,b] ⊂ (0,) → R is said to

be p-symmetric with respect to[

ap+bp

2

] 1p

if

f (t1p ) = f

([ap +bp− t]

1p

)holds for t ∈ [a,b].

1.4 FRACTIONAL CALCULUS 13

1.4 Fractional Calculus

Fractional calculus is a theory of differential and integral operators of non-integer order.In recent years considerable interest in this theory has been stimulated by the applicationsthat this calculus finds in numerical analysis and different areas of physics and engineer-ing. Fractional calculus made it possible to adopt a theoretical model on experimentaldata. There are several well known forms of the fractional operators (meaning fractionalintegral and fractional derivative) that have been studied extensively for their applications:Riemann-Liouville, Weyl, Erdely-Kober, Hadamard, Katugampola are just a few.

Fractional integrals and fractional derivatives will be observed in the real domain. Let[a,b]⊂R be a finite interval, that is −< a < b <. For the integral part of a real number we use notation [].

1.4.1 The Riemann-Liouville Fractional Integrals and Derivatives

More on the Riemann-Liouville fractional integrals and derivatives can be found in mono-graphs [21, 92, 135].

Definition 1.23 ([92]) Let > 0 and f ∈ L1[a,b]. The left-sided and the right-sidedRiemann-Liouville fractional integrals Ja+ f and Jb− f of order are defined by

Ja+ f (x) =1

()

∫ x

a(x− t)−1 f (t)dt , x ∈ [a,b] , (1.23)

Jb− f (x) =1

()

∫ b

x(t − x)−1 f (t)dt , x ∈ [a,b] . (1.24)

For = n ∈ N fractional integrals are actually n-fold integrals, that is

Jna+ f (x) =

∫ x

adt1

∫ t1

adt2 · · ·

∫ tn−1

af (tn)dtn

=1

(n−1)!

∫ x

a(x− t)n−1 f (t)dt , (1.25)

Jnb− f (x) =

∫ b

xdt1

∫ b

t1dt2 · · ·

∫ b

tn−1

f (tn)dtn

=1

(n−1)!

∫ b

x(t − x)n−1 f (t)dt . (1.26)

Lemma 1.4 Let , > 0 and f ∈ Lp[a,b], 1 ≤ p ≤ . Then for almost every x ∈ [a,b]we have

Ja+Ja+ f (x) = J+a+ f (x) , Jb−Jb− f (x) = J+

b− f (x) . (1.27)

If f ∈C[a,b] or + > 1, then equalities (1.27) hold for each x in [a,b].

14 1 PRELIMINARIES

The homogeneous Abel integral equation has only trivial solution (see Samko et al.[135], Section 2.4).

Lemma 1.5 Let > 0 and f ∈ L1[a,b]. Then integral equations Ja+ f = 0 and Jb− f = 0have only trivial solution f = 0 (almost everywhere).

Next result shows that the Riemann-Liouville fractional integral is bounded operatoron Lp[a,b].

Lemma 1.6 Let > 0 and 1 ≤ p ≤ . Then the Riemann-Liouville fractional integralsare bounded on Lp[a,b], that is

||Ja+ f ||p ≤ K|| f ||p , ||Jb− f ||p ≤ K|| f ||p , (1.28)

where

K =(b−a)

( +1).

For C[a,b] we have

||Ja+ f ||C ≤ K|| f ||C , ||Jb− f ||C ≤ K|| f ||C . (1.29)

We continue with definition and properties of the Riemann-Liouville fractional deriva-tives.

Definition 1.24 Let > 0, n = []+1 and f : [a,b] → R. The left-sided and the right-sided Riemann-Liouville fractional derivatives D

a+ f and Db− f of order are defined by

Da+ f (x) =

dn

dxn Jn−a+ f (x)

=1

(n−)dn

dxn

∫ x

a(x− t)n−−1 f (t)dt , (1.30)

Db− f (x) = (−1)n dn

dxn Jn−b− f (x)

=(−1)n

(n−)dn

dxn

∫ b

x(t− x)n−−1 f (t)dt . (1.31)

In particular, if 0 < < 1, then

Da+ f (x) =

1(1−)

ddx

∫ x

a(x− t)− f (t)dt , (1.32)

Db− f (x) =

−1(1−)

ddx

∫ b

x(t− x)− f (t)dt . (1.33)

For = n ∈ N we have

Dna+ f (x) = f (n)(x) , Dn

b− f (x) = (−1)n f (n)(x) , (1.34)

1.4 FRACTIONAL CALCULUS 15

and for = 0D0

a+ f (x) = D0b− f (x) = f (x) . (1.35)

The following result indicates that the functions (x− a)− j and (b− x)− j, play thesame role for the Riemann-Liouville fractional derivatives as the constants do in usualdifferentiation.

Lemma 1.7 Let > 0 and n = []+1.

(i) The equality Da+ f (x) = 0 is valid if and only if f (x) =

nj=1

c j(x− a)− j, where

c j ∈ R ( j = 1, . . . ,n) are arbitrary constants.In particular, when 0 < ≤ 1, the relation D

a+ f (x) = 0 holds if and only iff (x) = c(x−a)−1 with every c ∈ R.

(ii) The equality Db− f (x) = 0 is valid if and only if f (x) =

nj=1

d j(b− x)− j , where

d j ∈ R ( j = 1, . . . ,n) are arbitrary constants.In particular, when 0 < ≤ 1, the relation D

b− f (x) = 0 holds if and only iff (x) = d(b− x)−1 with every d ∈ R.

We end with conditions for the existence of fractional derivatives in the space ACn[a,b].

Theorem 1.4 Let ≥ 0 and n = [] + 1. If f ∈ ACn[a,b], then the Riemann-Liouvillefractional derivatives D

a+ f and Db− f exist almost everywhere on [a,b] and can be repre-

sented in the forms

Da+ f (x) =

n−1

k=0

f (k)(a)(k−+1)

(x−a)k− +1

(n−)

∫ x

a(x− t)n−−1 f (n)(t)dt , (1.36)

Db− f (x) =

n−1

k=0

(−1)k f (k)(b)(k−+1)

(b− x)k− +(−1)n

(n−)

∫ b

x(t − x)n−−1 f (n)(t)dt . (1.37)

Chapter2An Extended GeneralizedMittag-Leffler Function

The Mittag-Leffler function of one parameter is defined by the power series using thegamma function

E(z) =

n=0

zn

(n+1), (z, ∈ C ,() > 0) (2.1)

and it is a natural extension of the exponential, hyperbolic and trigonometric functions.For instance:

E0(z) =

n=0

zn =1

1− z,

E1(z) =

n=0

zn

(n+1)=

n=0

zn

n!= exp(z),

E2(z) = cosh(√

z),

E3(z) =13

[exp

(z

13

)+2exp

(−z

13

2

)cos

(12

√3z

13

)],

E4(z) =12

[cos(z

14 )+ cosh(z

14 )],

E2(−z2) = cosz.

This function and its generalizations appear as solutions of fractional order differential orintegral equations. First is the well known Wiman’s generalization, also known as Mittag-

17

18 2 AN EXTENDED GENERALIZED MITTAG-LEFFLER FUNCTION

Leffler function of two parameters:

E , (z) =

n=0

zn

(n+), (z, , ∈ C ,() > 0) , (2.2)

followed by Prabhakar’s function [123], i.e. Mittag-Leffler function of three parameters:

E , (z) =

n=0

( )n

(n+)zn

n!, (z, , , ∈ C ,() > 0) . (2.3)

Next extension was introduced by Shukla and Prajapati in [143]:

E ,q , (z) =

n=0

( )nq

(n+)zn

n!, (2.4)

where z, , , ,q ∈ C, () > max{0,(q)−1},(q) > 0, and ( )nq denotes the gener-alized Pochhammer symbol (1.6).Further, Salim and Faraj presented in [134] the function

E ,q,r , ,(z) =

n=0

( )nq

(n+)zn

()nr, (2.5)

where z, , , , ∈ C, min{(),(),( ),()} > 0, q,r > 0 and q ≤ () + r.Another recent extension is defined by Rahman et al. in [127]

E ,q,c , (z; p) =

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)zn

n!, (2.6)

where z, , , ,c ∈ C, (),( ),(c) > 0, p ≥ 0, q > 0 with Bp as an extension of thebeta function (1.9).

This chapter is based on our results from [15].

2.1 A Further Extension of the Mittag-Leffler Function

We define more extended and generalized Mittag-Leffler function E ,c,q,r , , (z; p) as follows:

Definition 2.1 Let , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0 with p ≥0, r > 0 and 0 < q ≤ r +(). Then the extended generalized Mittag-Leffler function

E ,c,q,r , , (z; p) is defined by

E ,c,q,r , , (z; p) =

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)zn

()nr. (2.7)

For different parameter choices, the correspondingknown generalizations of Mittag-Lefflerfunction can be deduced, as described below.

2.1 A FURTHER EXTENSION OF THE MITTAG-LEFFLER FUNCTION 19

Remark 2.1 Evidently, (2.7) is a generalization of the following functions:

(i) setting p = 0, it reduces to (2.5), which is the Salim-Faraj function E ,,q , ,r(z) defined

in [134],

(ii) setting = r = 1, it reduces to (2.6), which is the function E ,q,c , (z; p) defined by

Rahman et al. in [127],

(iii) setting p = 0 and = r = 1, it reduces to (2.4), which is the Shukla-Prajapati function

E ,q , (z) defined in [143],

(iv) setting p = 0 and = r = q = 1, it reduces to (2.3), which is the Prabhakar functionE , (z) defined in [123],

(v) setting p = 0 and = r = q = = 1, it reduces to (2.2), i.e. the Wiman functionE , (z), which for = 1 results with the Mittag-Leffler function E(z), the equation(2.1).

Theorem 2.1 The series in (2.7) is absolutely convergent for all values of z provided that

q < r+(). Moreover, if q = r+(), then E ,c,q,r , , (z; p) converges for |z| < rr ()()

qq .

Proof. Let E ,c,q,r , , (z; p) =

n=0

anzn, where

an =Bp( +nq,c− )

B( ,c− )(c)nq

(n+)()nr.

Using the Cauchy-Hadamard formula for the radius of convergence

R =1

limsupn→

n√|an|

and its alternative formula

R = limsupn→

∣∣∣∣ an

an+1

∣∣∣∣ ,the asymptotic formula for gamma function

(z+a)(z+b)

= za−b[1+

(a−b)(a+b−1)2z

+O

(1z2

)], |z| → , |argz| <

and the asymptotic behaviour of the generalized beta function for large x (with y and pfinite) by Chaudhry et al. ([30])

Bp(x,y) ∼ 12 p

14

x34

( px

) y−12

exp

[−2(px)

12 − 3p

2

](x → ),

we obtain limn→

an = 0 and


∣∣∣∣ an

an+1

∣∣∣∣ =∣∣∣∣ Bp( +nq,c− )Bp( +(n+1)q,c− )

· (c)nq

(c)(n+1)q· ()(n+1)r

()nr· ((n+1)+)

(n+)

∣∣∣∣∼(

1+q

+nq

) 2(c− )+14

exp[2√

p( +nq+q)−2√

p( +nq)]

×(nq)−q[1+

−q(2c+q−1)2nq

+O

(1

(nq)2

)]

×(nr)r[1+

r(2 + r−1)2nr

+O

(1

(nr)2

)]

×(n)[1+

(2 +−1)2n

+O

(1

(n)2

)], n → .

Hence,

limn→

∣∣∣∣ an

an+1

∣∣∣∣ = limn→

rr

qq nr+−q = R.

This means that the function E ,c,q,r , , (z; p) converges for all z provided that q < r +(),

and it is an entire function. Moreover if q = r +(), then E ,c,q,r , , (z; p) converges for

|z| < rr ()()

qq . �

The following theorems list some basic properties of this function. First we have somerecurrence relations.

Theorem 2.2 If , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0 with p ≥ 0,r > 0 and 0 < q < r+(), then

E ,c,q,r , , (z; p)−E ,c,q,r

, ,−1(z; p) =zr

1− ddz

E ,c,q,r , , (z; p), () > 1; (2.8)

E ,c,q,r , , (z; p) = E ,c,q,r

,+1,(z; p)+zddz

E ,c,q,r ,+1,(z; p). (2.9)

Proof.

E ,c,q,r , , (z; p)−E ,c,q,r

, ,−1(z; p) =

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)

[1

()nr− 1

(−1)nr

]zn

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)()

( +nr)

[nr

1−

]zn

=zr

1−

n=1

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)nzn−1

()nr

=zr

1− ddz

E ,c,q,r , , (z; p)

which proves (2.8).

2.1 A FURTHER EXTENSION OF THE MITTAG-LEFFLER FUNCTION 21

Further,

E ,c,q,r , , (z; p) =

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)zn

()nr

= (n+)

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n++1)zn

()nr

= (n+)E ,c,q,r ,+1,(z; p)

= E ,c,q,r ,+1,(z; p)+z

ddz

E ,c,q,r ,+1,(z; p),

hence (2.9) is proved. �

Next are some differential relations.

Theorem 2.3 If m ∈ N, w, , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0with p ≥ 0, r > 0 and 0 < q < r+(), then(

ddz

)m

E ,c,q,r , , (z; p) =

(c)mq

()mr

n=0

Bp( +(n+m)q,c− )B( ,c− )

(c+mq)nq

((n+m)+)(n+1)mzn

( +mr)nr;

(2.10)(ddz

)m [z−1E ,c,q,r

, , (wz ; p)]

= z−m−1E ,c,q,r ,−m,(wz ; p), () > m. (2.11)

Proof. We have(ddz

)m

E ,c,q,r , , (z; p)

=

n=m

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)[n(n−1) · · ·(n− (m−1))]zn−m

()nr

=

n=0

Bp( +(n+m)q,c− )B( ,c− )

(c)(n+m)q

((n+m)+)

× [(n+m)(n+m−1) · · ·(n+1)]zn

()(n+m)r.

From (x)a+b = (x+a)b(x)a follows (2.10).Further, (

ddz

)m [z−1E ,c,q,r

, , (wz ; p)]

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)

× [(n+−1)(n+−2) · · ·(n+−m)]wnzn+−m−1

()nr

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+−m)wnzn+−m−1

()nr.

�


Finally, we observe some special cases of the extended generalizedMittag-Leffler func-tion that we will use to prove our inequalities in the forthcoming chapters.

• if z = 0, then

E ,c,q,r , , (0; p) =

Bp( ,c− )B( ,c− )

1()

,

• if p = z = 0, then

E ,c,q,r , , (0;0) =

1()

.

2.2 Fractional Integral Operators Associatedwith the Mittag-Leffler Function

The corresponding generalized fractional integral operators, the left-sided w, ,c,q,ra+, , , f and

the right-sided w, ,c,q,rb−, , , f , which contain the extended generalized Mittag-Leffler function

as its kernel, we define by:

Definition 2.2 Let w, , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0 withp ≥ 0, r > 0 and 0 < q ≤ r +(). Let f ∈ L1[a,b] and x ∈ [a,b]. Then the left-sided

and the right-sided generalized fractional integral operator w, ,c,q,ra+, , , f and w, ,c,q,r

b−, , , f aredefined by (

w, ,c,q,ra+, , , f

)(x; p) =

∫ x

a(x− t)−1E ,c,q,r

, , (w(x− t) ; p) f (t)dt, (2.12)

(w, ,c,q,rb−, , , f

)(x; p) =

∫ b

x(t− x)−1E ,c,q,r

, , (w(t − x) ; p) f (t)dt. (2.13)

If we apply different parameter choices, then the corresponding known generalizationsof fractional integral operators can be deduced. We list those associated with the left-sidedfractional integral operator. Analogously holds for the right-sided.

Remark 2.2 (2.12) is a generalization of the following fractional integral operators:

(i) setting p = 0, it reduces to the Salim-Faraj fractional integral operator w, ,q,ra+, , , f (x)

defined in [134],

(ii) setting = r = 1, it reduces to the fractional integral operator w, ,q,ca+, , f (x; p) defined

by Rahman et al. in [127],

(iii) setting p = 0 and = r = 1, it reduces to the Srivastava-Tomovski fractional integral

operator w, ,qa+, , f (x) defined in [144],

2.2 OPERATORS WITH THE MITTAG-LEFFLER FUNCTION 23

(iv) setting p = 0 and = r = q = 1, it reduces to the Prabhakar fractional integraloperator ( , ; ;w) f (x) defined in [123],

(v) setting p = w = 0, it reduces to the left-sided Riemann-Liouville fractional integralJa+ f of order as in (1.23).

We follow with some properties of these operators.

Theorem 2.4 If w, , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0, p ≥ 0,r > 0, 0 < q ≤ r +() with f ∈ L1[a,b], x ∈ [a,b], then the left-sided fractional integral

operator w, ,c,q,ra+, , , and the right-sided w, ,c,q,r

b−, , , are bounded on L1[a,b] and

∥∥∥w, ,c,q,ra+, , , f

∥∥∥1≤C‖ f‖1 (2.14)

and ∥∥∥w, ,c,q,rb−, , , f

∥∥∥1≤C‖ f‖1 , (2.15)

where the constant C (0 < C < ) is given by

C=(b−a)()

n=0

∣∣Bp(+nq,c− )∣∣

|B( ,c− )|

∣∣(c)nq∣∣

(()n+()) |(n+)|

∣∣∣w(b−a)()∣∣∣n

|()nr| .

(2.16)

Proof. Using Fubini’s theorem we obtain

∥∥∥w, ,c,q,ra+, , , f

∥∥∥1

=∫ b

a

∣∣∣∣∫ x


, , (w(x− t) ; p) f (t)dt

∣∣∣∣dx

≤∫ b

a| f (t)|

[∫ b

t(x− t)()−1

∣∣∣E ,c,q,r , , (w(x− t) ; p)

∣∣∣dx

]dt

=∫ b

a| f (t)|

[∫ b−t

0u()−1E ,c,q,r

, , (wu ; p)|du

]dt

≤∫ b

a| f (t)|

[∫ b−a

0u()−1E ,c,q,r

, , (wu ; p)|du

]dt

≤

n=0

∣∣Bp( +nq,c− )∣∣

|B( ,c− )|

∣∣(c)nq∣∣

|(n+)||wn||()nr|

×∫ b−a

0u()n+()−1du‖ f‖1

= C‖ f‖1 .

We can see that the constant C, defined by (2.16), is finite.For the right-sided fractional integral operator, inequality (2.15) can be proved analogously.

�

Next we have several results for a fractional integral operator applied on a power func-tion.


Theorem 2.5 If ,w, , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0 withp ≥ 0, r > 0 and 0 < q ≤ r+(), then for f (t) = (t −a)−1 follows

(w, ,c,q,ra+, , ,(t−a)−1

)(x; p) = ()(x−a)+−1E ,c,q,r

,+,(w(x−a) ; p). (2.17)

Proof. By definition for the left-sided fractional integral operator, we have(w, ,c,q,ra+, , ,(t−a)−1

)(x; p)

=∫ x


, , (w(x− t) ; p)(t −a)−1dt

=∫ x

a(x− t)−1

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn(x− t)n

()nr(t−a)−1dt

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn

()nr

∫ x

a(x− t)n+−1(t −a)−1dt

Now we use the substitution t = x− s(x−a), as in Example 1.1, to obtain

∫ x

a(x− t)n+−1(t−a)−1dt = (x−a)n++−1B( ,n+),

from which follows(w, ,c,q,ra+, , ,(t−a)−1

)(x; p)

= (x−a)+−1

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn

()nr(x−a)n()(n+)

(n+ + )

= ()(x−a)+−1E ,c,q,r ,+,(w(x−a) ; p).

�

If we set a = 0 and x = 1 in the previous theorem, then we obtain a following result.

Corollary 2.1 If ,w, , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0 withp ≥ 0, r > 0 and 0 < q ≤ r+(), then

1()

∫ 1

0t−1(1− t)−1E ,c,q,r

, , (w(1− t) ; p)dt = E ,c,q,r ,+,(w; p).

Setting = 1, we get following identity for the constant function.

Corollary 2.2 If w, , ,, ,c∈C,(),(),() > 0,(c) >( ) > 0 with p≥ 0,r > 0 and 0 < q ≤ r+(), then

(w, ,c,q,ra+, , ,1

)(x; p) = (x−a)E ,c,q,r

,+1,(w(x−a) ; p). (2.18)

2.2 OPERATORS WITH THE MITTAG-LEFFLER FUNCTION 25

Proof. Identity (2.18) is a direct consequence of the Theorem 2.5, but it can also be easilyobtain as follows:(

w, ,c,q,ra+, , ,1

)(x; p) =

∫ x


, , (w(x− t) ; p)dt

=∫ x

a(x− t)−1

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn(x− t)n

()ndt

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn

()n

∫ x

a(x− t)n+−1dt

= (x−a)

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn

()n

(x−a)n

n+

= (x−a)E ,c,q,r ,+1,(w(x−a) ; p). �

Similarly, for the right-sided fractional integral operator follows:

Theorem 2.6 If ,w, , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0 withp ≥ 0, r > 0 and 0 < q ≤ r+(), then for g(t) = (b− t)−1 follows(

w, ,c,q,rb−, , ,(b− t)−1

)(x; p) = ()(b− x)+−1E ,c,q,r

,+,(w(b− x) ; p). (2.19)

Proof. By definition for the right-sided fractional integral operator, we have(w, ,c,q,rb−, , ,(b− t)−1

)(x; p)

=∫ b

x(t− x)−1E ,c,q,r

, , (w(t − x) ; p)(b− t)−1dt

=∫ b

x(t− x)−1

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn(t− x)n

()nr(b− t)−1dt

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn

()nr

∫ b

x(t − x)n+−1(b− t)−1dt

For the right-sided integral operator we use the substitution t = x+ s(b− x) (see Example1.1), to obtain∫ b

x(t− x)n+−1(b− t)−1dt = (b− x)n++−1B( ,n+),

from which we get(w, ,c,q,rb−, , ,(b− t)−1

)(x; p)

= (b− x)+−1

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)wn

()nr(b− x)n()(n+)

(n+ + )

= ()(b− x)+−1E ,c,q,r ,+,(w(b− x) ; p).

�

Setting b = 1 and x = 0 in previous theorem we have the next result.


Corollary 2.3 If ,w, , ,, ,c ∈ C, (),(),() > 0, (c) > ( ) > 0 withp ≥ 0, r > 0 and 0 < q ≤ r+(), then

1()

∫ 1

0t−1(1− t)−1E ,c,q,r

, , (wt ; p)dt = E ,c,q,r ,+,(w; p).

For the constant function, we obtain following identity.

Corollary 2.4 If w, , ,, ,c∈C,(),(),() > 0,(c) >( ) > 0 with p≥ 0,r > 0 and 0 < q ≤ r+(), then(

w, ,c,q,rb−, , ,1

)(x; p) = (b− x)E ,c,q,r

,+1,(w(b− x) ; p). (2.20)

2.3 Unified Integral Operators

In recent years considerable interest in fractional calculus has been stimulated by theapplications that this calculus finds in numerical analysis and different areas of physicsand engineering. All forms of the fractional operators that have been studied extensivelyfor their applications, in a special case are reduced to the left-sided and the right-sidedRiemann-Liouville fractional integrals Ja+ f and Jb− f of order defined as in [92, 135]for f ∈ L1[a,b] with (1.23) and (1.24).

Further generalization of the fractional integral operator we give in [46, 101] as follows:

Definition 2.3 [46] Let w, ,, ,c ∈ C, (),() > 0, (c) > ( ) > 0 with p ≥ 0, ,r > 0 and 0 < q ≤ r + . Let f ∈ L1[a,b], 0 < a < b < , be a positive function. Leth : [a,b] → R be a differentiable function, strictly increasing. Also let

x be an increasingfunction on [a,) and x ∈ [a,b]. Then the left and the right generalized fractional integral

operators h Fw, ,c,q,r

a+, , , f and h Fw, ,c,q,r

b−, , , f are defined by

(h Fw, ,c,q,r

a+, , , f)

(x; p)

=∫ x

a

(h(x)−h(t))h(x)−h(t)

E ,c,q,r , , (w(h(x)−h(t)) ; p)h′(t) f (t)dt (2.21)

and (h Fw, ,c,q,r

b−, , , f)

(x; p)

=∫ b

x

(h(t)−h(x))h(t)−h(x)

E ,c,q,r , , (w(h(t)−h(x)) ; p)h′(t) f (t)dt (2.22)

Next is a special case of the above operator, setting (x) = x , > 0. This operator

we denote hw, ,c,q,ra+, , , f (analogously h

w, ,c,q,rb−, , , f ):

2.3 UNIFIED INTEGRAL OPERATORS 27

Definition 2.4 [101] Let w, ,, ,c ∈C, (),() > 0, (c) >( ) > 0 with p≥ 0, ,r > 0 and 0 < q ≤ r + . Let f ∈ L1[a,b], 0 < a < b < , be a positive function. Leth : [a,b] → R be a differentiable function, strictly increasing and let x ∈ [a,b]. Then the

left and the right generalized fractional integral operators hw, ,c,q,ra+, , , f and h

w, ,c,q,rb−, , , f are

defined by (h

w, ,c,q,ra+, , , f

)(x; p)

=∫ x

a(h(x)−h(t))−1E ,c,q,r

, , (w(h(x)−h(t)) ; p)h′(t) f (t)dt (2.23)

and (h

w, ,c,q,rb−, , , f

)(x; p)

=∫ b

x(h(t)−h(x))−1E ,c,q,r

, , (w(h(t)−h(x)) ; p)h′(t) f (t)dt. (2.24)

If we set p = = 0 in this definition, then (2.23) reduces to the left-sided Riemann-Liouville fractional integral of a function f with respect to another function h of order ([92, 135]):

Ja+;h f (x) =1

()

∫ x

a(h(x)−h(t))−1h′(t) f (t)dt, x ∈ (a,b].

For the constant function we have following identities.

Proposition 2.1 Let w, ,, ,c ∈ C, (),() > 0, (c) > ( ) > 0 with p ≥ 0, ,r > 0 and 0 < q ≤ r + . Let 0 < a < b < and let h : [a,b] → R be a differentiablefunction, strictly increasing. Then(

hw, ,c,q,ra+ , , ,1

)(x; p) = (h(x)−h(a))E ,c,q,r

,+1,(w(h(x)−h(a)) ; p)

and (h

w, ,c,q,rb−, , ,1

)(x; p) = (h(b)−h(x))E ,c,q,r

,+1,(w(h(b)−h(x)) ; p).

Proof. Analogously as in Corollary 2.2. �

Chapter3Opial Type Fractional IntegralInequalities Associated withthe Mittag-Leffler Function

Inequalities which involve integrals of functions and their derivatives, whose study has ahistory of about one century, are of great importance in mathematics, with far-reachingapplications in the theory of differential equations, approximations and probability, amongothers. They occupy a central position in mathematical analysis and its applications.

In 1960. Opial published an inequality involving integrals of a function and its deriva-tive, which now bears his name ([117]):

Theorem 3.1 (OPIAL’S INEQUALITY) Let f ∈C1[0,h] be such that f (0) = f (h) = 0 andf (x) > 0 for x ∈ (0,h). then∫ h

0

∣∣ f (x) f ′(x)∣∣dx ≤ h

4

∫ h

0

[f ′(x)

]2dx , (3.1)

where constant h/4 is the best possible.

A monograph by Agarwal and Pang [6] is dedicated to the theory of Opial type inequal-ities and its applications. We also did several papers involving various extensions, gener-alizations and discrete analogues (see [8]-[25], [60]-[63]). To present Mitrinovic-Pecaricgeneralizations of Opial type inequalities given in [121], we need the next characteriza-tion: We say that a function u : [a,b] −→ R belongs to the class U(v,K) if it admits therepresentation

u(x) =∫ x

aK(x,t)v(t)dt ,

29

30 3 OPIAL TYPE FRACTIONAL INTEGRAL INEQUALITIES

where v is a continuous function and K is an arbitrary nonnegative kernel such that v(x) > 0implies u(x) > 0 for every x ∈ [a,b]. We also assume that all integrals under considerationexist and converge. In the following theorems we have Opial type inequalities.

Theorem 3.2 ([121, p. 236]) Let u1 ∈U(v1,K), u2 ∈U(v2,K) and v2(x) > 0 for everyx ∈ [a,b]. Further, let (u) and f (u) be convex and increasing for u ≥ 0 and f (0) = 0. Iff is a differentiable function and M = max K(x,t), then

M

b∫a

v2 (t)(∣∣∣∣v1 (t)

v2 (t)

∣∣∣∣)

f ′(

u2 (t)(∣∣∣∣u1 (t)

u2 (t)

∣∣∣∣))

dt

≤ f

⎛⎝M

b∫a

v2 (t)(∣∣∣∣v1 (t)

v2 (t)

∣∣∣∣)

dt

⎞⎠ .

Theorem 3.3 ([121, p. 238])Let : [0,)−→R be a differentiable function such that for

m > 1 the function (x1m ) is convex and (0)= 0. Let u∈U(v,K) where

(∫ xa (K(x,t))l dt

) 1l ≤

M, l−1 +m−1 = 1. Then

∫ b

a| u(x) |1−m ′(| u(x) |) | v(x) |m dx ≤ m

Mm

(M

(∫ b

a| v(x) |m dx

) 1m)

.

If the function (x1m ) is concave, then the reverse inequality holds.

In [8, 10], also [21], we gave the following generalizations of the above inequalities.

Theorem 3.4 ([21, p. 86]) Let u1 ∈ U(v1,K), u2 ∈ U(v2,K) and v2(x) > 0 for everyx ∈ [a,b]. Further, let (u) be convex, nonnegative and increasing for u ≥ 0, f (u) beconvex for u ≥ 0 and f (0) = 0. If f is a differentiable function and M = max K(x,t), then

M∫ b

av2(t)

(∣∣∣∣v1(t)v2(t)

∣∣∣∣)

f ′(

u2(t)(∣∣∣∣u1(t)

u2(t)

∣∣∣∣))

dt

≤ f

(M∫ b

av2(t)

(∣∣∣∣v1(t)v2(t)

∣∣∣∣)

dt

)

≤ 1b−a

∫ b

af

(M (b−a)v2(t)

(∣∣∣∣v1(t)v2(t)

∣∣∣∣))

dt.

Theorem 3.5 ([21, p. 101]) Let : [0,) → R be a differentiable function such that for

m > 1 the function (x1m ) is convex and (0)= 0. Let u∈U(v,K), where (

∫ xa (K(x,t))ldt)

1l ≤

M and l−1 +m−1 = 1. Then

∫ b

a

∣∣u(x)∣∣1−m ′ (∣∣u(x)

∣∣)∣∣v(x)∣∣mdx ≤ mMm

(M

(∫ b

a

∣∣v(x)∣∣mdx

) 1m)

≤ mMm (b−a)

∫ b

a((b−a)

1m M

∣∣v(x∣∣)dx. (3.2)

If the function (x1m ) is concave, then the reverse inequalities hold.

3.1 OPIAL TYPE INEQUALITIES AND MITTAG-LEFFLER FUNCTION 31

Fractional differentiation inequalities have applications to fractional differential equa-tions; the most important ones are in establishing uniqueness of the solution of initialproblems and giving upper bounds to their solutions. These applications have motivatedmany researchers in the field of integral inequalities to explore certain extensions and gen-eralizations by using different fractional differential and integral operators.

The right-sided versions of following inequalities can be established and proved anal-ogously by using the right-sided fractional integral operator w, ,c,q,r

b−, , , f defined with (2.13).

This chapter is based on our results from [15].

3.1 Opial Type Fractional Integral Inequalities andan Extended Generalized Mittag-Leffler Function

In this section we use the extended generalized Mittag-Leffler function E ,c,q,r , , with the

corresponding fractional integral operator w, ,c,q,ra+, , , f (in real domain) to obtain fractional

generalizations of Opial type inequalities due to Mitrinovic and Pecaric.

Here, for the reader’s convenience we will use a simplified notation

EEE(z; p) := E ,c,q,r , , (z; p) (3.3)

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)zn

()nr,

( f )(x; p) :=(w, ,c,q,ra+, , , f

)(x; p) (3.4)

=∫ x

a(x− t)−1EEE(w(x− t) ; p) f (t)dt.

The first result is a generalization of Theorem 3.4. As mentioned before, from Theorem3.4 immediately follows Theorem 3.2.

Theorem 3.6 Let w∈R, r, , > 0, c > > 0 with p≥ 0 and 0 < q≤ r+ . Let u1,u2,and f be the same as in Theorem 3.4. Then for > 1 following inequalities hold

EEE(w(b−a) ; p)(b−a)−1

×∫ b

av2(x)

(∣∣∣∣v1(x)v2(x)

∣∣∣∣)

f ′(

(v2)(x; p)(∣∣∣∣ (v1)(x; p)

(v2)(x; p)

∣∣∣∣))

dx

≤ f

(EEE(w(b−a) ; p)(b−a)−1

∫ b

av2(x)

(∣∣∣∣v1(x)v2(x)

∣∣∣∣)

dx

)(3.5)

≤ 1b−a

∫ b

af

(EEE(w(b−a) ; p)(b−a)v2(x)

(∣∣∣∣v1(x)v2(x)

∣∣∣∣))

dx. (3.6)


Proof. Let us define the following kernel

K(x,t) ={

(x− t)−1EEE(w(x− t) ; p) a ≤ t ≤ x,0 x < t ≤ b,

(3.7)

and denote the extended generalized fractional integral operators v1 and v2 by u1 and u2

as

ui(x) = (vi)(x; p) =∫ x

a(x− t)−1EEE(w(x− t) ; p)vi(t)dt, (3.8)

for i = 1,2. Next we have

n=0

p( +n,c− )B( ,c− )

(c)nqwn(x− t)n

(n+)()nr≤

n=0

p( +n,c− )B( ,c− )

(c)nqwn(b−a)n

(n+)()nr

= EEE(w(b−a) ; p),

where convergence of EEE(w(b−a) ; p) follows by Theorem 2.1. Using

(x− t)−1 ≤ (x−a)−1 ≤ (b−a)−1, > 1,

we obtain

K(x, t) ≤EEE(w(b−a) ; p)(b−a)−1, > 1, q < r+().

Finally, if we setM = EEE(w(b−a) ; p)(b−a)−1

and use here the functions u1,u2, then respectively, by Theorem 3.4 inequalities (3.5) and(3.6) follow. �

Remark 3.1 For different choices of parameters we can obtain the corresponding frac-tional integral inequalities, such as:

(i) setting p = 0 in (3.5), we get [62, Theorem 2.2],

(ii) setting p = 0 in (3.6), we get [62, Theorem 2.3] (there is a misprint in the (2.5) in[62, Theorem 2.3]: instead of (b−a)−1 it should be (b−a) ),

(iii) setting p = 0 and = r = q = 1, we get Opial type inequalities for Prabhakar frac-tional integral operator defined in [123],

(iv) setting p = 0, = r = q = 1 and w = 0, we get the result for left-sided Riemann-Liouville fractional integral in [10, Corollary 3.2].

If we set (x) = xl+m in Theorem 3.6, then we have the following result.

Corollary 3.1 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let u1,u2and f be the same as in Theorem 3.4. Let m, l > 1 with l−1 +m−1 = 1. Then for > 1following inequalities hold

EEE(w(b−a) ; p)(b−a)−1

×∫ b

av2(x)

∣∣∣∣ v1(x)v2(x)

∣∣∣∣l+m

f ′(

(v2)(x; p)∣∣∣∣ (v1) (x; p)(v2) (x; p)

∣∣∣∣l+m

)dx

3.1 OPIAL TYPE INEQUALITIES AND MITTAG-LEFFLER FUNCTION 33

≤ f

(EEE(w(b−a) ; p)(b−a)−1

∫ b

av2(x)

∣∣∣∣v1(x)v2(x)

∣∣∣∣l+m

dx

)

≤ 1b−a

∫ b

af

(EEE(w(b−a) ; p)(b−a)v2(x)

∣∣∣∣v1(x)v2(x)

∣∣∣∣l+m

)dx.

The next result is a generalization of Theorem 3.5 from which immediately followsTheorem 3.3.

Theorem 3.7 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let l,m,uand be the same as in Theorem 3.5. Then for > 1

m following inequalities hold

∫ b

a|(v) (x; p)|1−m ′ (|(v)(x; p)|) |v(x)|m dx

≤ mlml( − 1

m

)ml

(EEE(w(b−a) ; p)m (b−a)m−1

×

⎛⎝EEE(w(b−a) ; p)

(b−a)−1m

l1l( − 1

m

) 1l

(∫ b

a|v(x)|m dx

) 1m

⎞⎠ (3.9)

≤ mlml( − 1

m

)ml

(EEE(w(b−a) ; p))m (b−a)m

×∫ b

a


(b−a)

l1l( − 1

m

) 1l

|v(x)|⎞⎠dx. (3.10)

Proof. Let us define the kernel K(x,t) as (3.7) and the extended generalized fractionalintegral operators v by u as (3.8). Using the same argument as in the proof of Theorem3.6 we obtain

EEE(w(x− t) ; p) ≤EEE(w(b−a) ; p)

from which follows(∫ x

a(K(x,t))l dt

) 1l

≤ EEE(w(b−a) ; p)(∫ x

a(x− t)l(−1) dt

) 1l

= EEE(w(b−a) ; p)(x−a)−

1m

l1l( − 1

m

) 1l

.

Since > 1m , if we use

M = EEE(w(b−a) ; p)(b−a)−

1m

l1l( − 1

m

) 1l

(3.11)

and u(x) = (v)(x; p), then by Theorem 3.5 inequalities (3.9) and (3.10) follow, respec-tively. �


Remark 3.2 Let the assumptions of Theorem 3.7 hold. For > 1 we have

∫ b

a|(v)(x; p)|1−m ′ (|(v)(x; p)|) |v(x)|m dx

≤ m

(EEE(w(b−a) ; p)m (b−a)m−1

×

(EEE(w(b−a) ; p)(b−a)−

1m

(∫ b

a|v(x)|m dx

) 1m)

≤ m


×∫ b

a (EEE(w(b−a) ; p)(b−a) |v(x)|)dx.

These inequalities will follow if we use the next estimate

K(x, t) ≤EEE(w(b−a) ; p)(b−a)−1, > 1, q < r+(),

and set M to be(∫ x

a(K(x,t))l dt

) 1l

≤ EEE(w(b−a) ; p)(b−a)−1(∫ x

adt

) 1l

≤ EEE(w(b−a) ; p)(b−a)−1m = M.

Setting p = 0, we get [62, Theorem 2.5], hence Theorem 3.7 is a generalization and animprovement of [62, Theorem 2.5].

Also, others corresponding fractional integral inequalities can be obtain by fixing cer-tain parameters in Theorem 3.7, such as:

(i) setting p = 0 and = r = q = 1, we get Opial type inequalities for Prabhakar frac-tional integral operator defined in [123],

(ii) setting p = 0, = r = q = 1 and w = 0, we get the result for left-sided Riemann-Liouville fractional integral in [8, Theorem 3.1].

If we consider (x) = xl+m in Theorem 3.7, then we have the following.

Corollary 3.2 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let l,mand u be the same as in Theorem 3.5. Then for > 1

m following inequalities hold

∫ b

a|(v)(x; p)|l |v(x)|m dx

≤ (EEE(w(b−a) ; p))l (b−a)l(−1m )

l2( − 1

m

) (∫ b

a|v(x)|m dx

)l

≤ (EEE(w(b−a) ; p))l (b−a)l

l2( − 1

m

) ∫ b

a|v(x)|l+m dx.

3.2 PROPERTIES OF ASSOCIATED FRACTIONAL LINEAR FUNCTIONAL 35

3.2 Properties of Associated FractionalLinear Functional

Motivated by the inequalities given in Theorems 3.4 and 3.5, we have studied functionalsderived from them, and we applied them with respect to several types of fractional inte-grals and fractional derivatives (see [9, 10], also [21]). Here we consider one of them, anonnegative difference in (3.2) define as follows

(u,v) =m

Mm (b−a)

∫ b

a((b−a)

1m M

∣∣v(x∣∣)dx

−∫ b

a

∣∣u(x)∣∣1−m ′ (∣∣u(x)

∣∣)∣∣v(x)∣∣mdx.

For this functional we have the following mean value theorems (given in [9], also [21]):

Theorem 3.8 ([21, p. 103]) Let : [0,) → R be a differentiable function such that for

m > 1 the function (x1m ) is convex and (0)= 0. Let u∈U(v,K) where (

∫ xa (K(x,t))ldt)

1l ≤

M and l−1 +m−1 = 1. If ∈ C2(I), where I ⊆ [0,) is closed interval, then there exists ∈ I such that the following equality holds

(u,v) = ′′

( )− (m−1) ′( )2m 2m−1

×(

(b−a)Mm∫ b

a

∣∣v(x)∣∣2mdx−2

∫ b

a

∣∣u(x)∣∣m∣∣v(x)∣∣mdx

).

Theorem 3.9 ([21, p. 104]) Let 1,2 : [0,) → R be differentiable functions such that

for m > 1 the function i(x1m ) is convex and i(0) = 0, i = 1,2. Let u ∈ U(v,K), where

(∫ xa (K(x, t))l dt)

1l ≤ M and l−1 + m−1 = 1. If 1,2 ∈ C2(I), where I ⊆ [0,) is closed

interval and

(b−a)Mm∫ b

a|v(x)|2m dx−2

∫ b

a|u(x)|q |v(x)|m dx �= 0,

then there exists ∈ I such that we have

1(u,v)2(u,v)

= ′′

1 ( )− (m−1) ′1( )

′′2 ( )− (m−1) ′

2( ),

provided that denominators are not equal to zero.

Hence, we study a functional derived from the inequality (3.10), i.e. its nonnegativedifference, which we denote by :

(u,v) =ml

ml(− 1

m

)ml

(EEE(w(b−a) ; p))m (b−a)m ×∫ b

a


(b−a)

l1l(− 1

m

) 1l

|v(x)|⎞⎠dx

−∫ b

a|(v)(x; p)|1−m ′ (|(v) (x; p)|) |v(x)|m dx.


A simplified notation EEE and f as in (3.3) and (3.4) is used. For this functional wepresent following mean value theorems.

Theorem 3.10 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let

: [0,)→R be a differentiable function such that for m > 1 the function (x1m ) is convex

and (0) = 0. Let l−1 +m−1 = 1, > 1m and v ∈ L1[a,b]. If ∈C2(I), where I ⊆ [0,) is

closed interval, then there exists ∈ I such that the following equality holds

((v) (x; p),v(x)) = ′′

( )− (m−1) ′( )2m 2m−1

×[


lml( − 1

m

)ml

∫ b

a|v(x)|2m dx

−2∫ b

a|(v)(x; p)|m |v(x)|m dx

]. (3.12)

Proof. It follows directly for M defined by (3.11), function u(x) = (v)(x; p) and Theo-rem 3.8. �

Theorem 3.11 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let

1,2 : [0,) → R be differentiable functions such that for m > 1 the function i(x1m ) is

convex and i(0) = 0, i = 1,2. Let l−1 +m−1 = 1, > 1m and v∈ L1[a,b]. If 1,2 ∈C2(I),

where I ⊆ [0,) is closed interval, then there exists ∈ I such that the following equalityholds

1 ((v)(x; p),v(x))2 (((v) (x; p),v(x))

= ′′

1 ( )− (m−1) ′1( )

′′2 ( )− (m−1) ′

2( ), (3.13)

provided that denominators are not equal to zero.

Proof. It follows directly for the function u(x) = (v)(x; p) and Theorem 3.9. �

Remark 3.3 For different choices of parameters we can get corresponding fractional in-tegral inequalities, such as:

(i) setting p = 0 in (3.12) and (3.13), using the same arguments as in Remark 3.2 weget improvements of Theorem 2.11 and Theorem 2.12 in [62], respectively,

(ii) setting p = 0, = r = q = 1 and w = 0 in (3.12) and (3.13), we get results for left-sided Riemann-Liouville fractional integral in Theorem 4.1 and Theorem 4.2 in [9],respectively.

Chapter4Polya-Szego and ChebyshevTypes Fractional IntegralInequalities Associated withthe Mittag-Leffler Function

The Chebyshev functional T ( f ,g) for two Lebesgue integrable functions f and g on inter-val [a,b] is defined by

T ( f ,g) =1

b−a

∫ b

af (x)g(x)dx−

(1

b−a

∫ b

af (x)dx

)(1

b−a

∫ b

ag(x)dx

). (4.1)

Majority of problems involving Chebyshev functional are to give a lower bound or anupper bound for T , under various assumptions. For instance, if f and g are monotonic inthe same sense (in the opposite sense) then we obtain a well-known Chebyshev inequality([32])

T ( f ,g) ≥ 0 (≤ 0). (4.2)

Also, if we have constants m,M,n,N ∈ R such that for x ∈ [a,b]

m ≤ f (x) ≤ M, n ≤ g(x) ≤ N,

then the Gruss inequality ([68]) states

|T ( f ,g)| ≤ (M−m)(N−n)4

. (4.3)

37

38 4 POLYA-SZEGO AND CHEBYSHEV TYPES INEQUALITIES

For more recent inequalities see [28, 39, 41, 84, 114, 115, 120]. Following inequalities arethe subject of our research: the first one was introduced by Polya and Szego ([122])(∫ b

a f 2(x)dx)(∫ b

a g2(x)dx)

(∫ ba f (x)g(x)dx

)2 ≤ 14

(√mnMN

+

√MNmn

)2

. (4.4)

Next is the inequality by Dragomir and Diamond ([43])

|T ( f ,g)| ≤ (M−m)(N−n)4√

mMnN

1

(b−a)2

∫ b

af (x)dx

∫ b

ag(x)dx. (4.5)

Using Karamata’s estimations of the Chebyshev quotient ([89]), Pecaric and Peric givegeneralized and improved inequality of (4.5) for positive normalized functional in ([120])

− (M−m)(N−n)(√mN +

√Mn

)2( f g) ≤ − (M−m)(N−n)(√mn+

√MN

)2( f )(g) ≤ ( f g)−( f )(g)

≤ (M−m)(N−n)(√mn+

√MN

)2( f g) ≤ (M−m)(N−n)(√

mN +√

Mn)2( f )(9). (4.6)

Motivated by the paper [114], where authors have proved Polya-Szego and Chebyshevtypes fractional integral inequalities for the Riemann-Liouville fractional integral operator,we presents improved and generalized corresponding results using our extended general-ized Mittag-Leffler function with its fractional integral operator.

The right-sided versions of following inequalities can be established and proved anal-ogously by using the right-sided fractional integral operator w, ,c,q,r

b−, , , f defined with (2.13).This chapter is based on our results from [14].

4.1 Polya-Szego Type Fractional IntegralInequalities and an ExtendedGeneralized Mittag-Leffler Function

In this section we use extended generalized Mittag-Leffler function E ,c,q,r , , with the corre-

sponding fractional integral operator w, ,c,q,ra+, , , f (in real domain) to obtain fractional gener-

alizations of inequalities due to Polya and Szego. Following theorems are based on [114]where this was done for the Riemann-Liouville fractional integral operator. The role of theparameter > 0 will be of great significance and for the reader’s convenience we will usea simplified notation

EEE (z; p) := E ,c,q,r , , (z; p) (4.7)

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)zn

()nr,

4.1 POLYA-SZEGO TYPE INEQUALITIES AND MITTAG-LEFFLER FUNCTION 39

( f )(x; p) :=(w, ,c,q,ra+, , , f

)(x; p) (4.8)

=∫ x

a(x− t)−1EEE (w(x− t) ; p) f (t)dt.

Theorem 4.1 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,g,1,2,1 and 2 be positive integrable functions on [0,) satisfying

0 < 1(u) ≤ f (u) ≤ 2(u), 0 < 1(u) ≤ g(u)≤ 2(u), u ∈ [a,x]. (4.9)

Then the following inequality holds(12 f 2

)(x; p)

(12g2

)(x; p)

[( (11 +22) f g) (x; p)]2≤ 1

4. (4.10)

Proof. From the given conditions follows(2(u)1(u)

− f (u)g(u)

)(f (u)g(u)

− 1(u)2(u)

)≥ 0,

that is

(1(u)1(u)+2(u)2(u)) f (u)g(u) ≥ 1(u)2(u)( f (u))2 +1(u)2(u)(g(u))2.

Multiplying above inequality by (x− u)−1EEE (w(x− u) ; p) and integrating on [a,x] weobtain ∫ x

a(x−u)−1EEE (w(x−u) ; p)(1(u)1(u)+2(u)2(u)) f (u)g(u)du

≥∫ x

a(x−u)−1EEE (w(x−u) ; p)1(u)2(u)( f (u))2du

+∫ x

a(x−u)−1EEE (w(x−u) ; p)1(u)2(u)(g(u))2du,

that is

( (11 +22) f g) (x; p)≥ (

12 f 2) (x; p)+(12g

2)(x; p).

Since a+b≥ 2√

ab for a,b ∈ R (the AM-GM inequality), we have

( (11 +22) f g) (x; p)

≥ 2√

(12 f 2)(x; p)(12g2)(x; p),

which leads to the inequality (4.10) as required. �

Fixing the bounds on functions f and g we get the following special case of Theorem4.1.


Corollary 4.1 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let fand g be two positive integrable functions on [0,) satisfying

0 < m ≤ f (u) ≤ M < , 0 < n ≤ g(u) ≤ N < , u ∈ [a,x]. (4.11)

Then the following inequality holds

( f 2

)(x; p)

(g2

)(x; p)

[( f g)(x; p)]2≤ 1

4

(√mnMN

+

√MNmn

)2

.

Remark 4.1 Choosing particular values of parameters in Theorem 4.1, known Polya-Szego type inequalities for several fractional integral operators can be deduced (for moredetails on fractional integral operators see [15] and references therein). For example, set-ting w = p = 0 (and a = 0), we get Polya-Szego inequality for the Riemann-Liouvillefractional integral operator given in [114, Lemma 3.1].

Now we prove the next Polya-Szego type inequality.

Theorem 4.2 Suppose that the assumptions of Theorem 4.1 hold with > 0. Then

(12) (x; p)(12

)(x; p)

( f 2

)(x; p)

(g2

)(x; p)[

(1 f ) (x; p)(1g

)(x; p)+ (2 f ) (x; p)

(2g

)(x; p)

]2 ≤ 14. (4.12)

Proof. Under given conditions on f ,g and i,i(i = 1,2) in (4.9), for u,v ∈ [a,t] we have

2(u)1(v)

− f (u)g(v)

≥ 0 andf (u)g(v)

− 1(u)2(v)

≥ 0,

which imply (1(u)2(v)

+2(u)1(v)

)f (u)g(v)

≥ ( f (u))2

(g(v))2 +1(u)2(u)1(v)2(v)

that is

1(u) f (u)1(v)g(v)+2(u) f (u)2(v)g(v) ≥ 1(v)2(v)( f (u))2 +1(u)2(u)(g(v))2.

Multiplying above inequality by

(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)

and integrating, we obtain∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)

×EEE (w(x− v) ; p)1(u) f (u)1(v)g(v)dudv

+∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)

×EEE (w(x− v) ; p)2(u) f (u)2(v)g(v)dudv

4.1 POLYA-SZEGO TYPE INEQUALITIES AND MITTAG-LEFFLER FUNCTION 41

≥ ∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)

×EEE (w(x− v) ; p)1(v)2(v)( f (u))2dudv

+∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)

×EEE (w(x− v) ; p)1(u)2(u)(g(v))2dudv,

that is

(1 f ) (x; p)(1g

)(x; p)+ (2 f ) (x; p)

(2g(x; p)

)≥ (

f 2)(x; p)(12

)(x; p)+ (12)(x; p)

(g2) (x; p).

Now if we apply the AM-GM inequality we get

(1 f ) (x; p)(1g

)(x; p)+ (2 f ) (x; p)

(2g

)(x; p)

≥ 2√

( f 2)(x; p)(12

)(x; p)(12)(x; p)

(g2

)(x; p)

which leads to the inequality (4.12). �

In the results that follow, we need next equality from Corollary 2.2:

(1)(x; p) = (x−a)EEE+1(w(x−a) ; p). (4.13)

We continue with a special case of Theorem 4.2.

Corollary 4.2 Suppose that the assumptions of Corollary 4.1 hold with > 0. Then

(1)(x; p)(1

)(x; p)

( f 2

)(x; p)

(g2

)(x; p)[

( f ) (x; p)(g

)(x; p)

]2 ≤ 14

(√mnMN

+

√MNmn

)2

.

Theorem 4.3 Suppose that the assumptions of Theorem 4.1 hold with > 0. Then( f 2)(x; p)

(g2)(x; p)

≤ ( (2 f g/1)) (x; p)( (2 f g/1)

)(x; p). (4.14)

Proof. Under given conditions on f ,g and i,i(i = 1,2) in (4.9), for u,v ∈ [a,t] we have

2(u) f (u)g(u)1(u)

− ( f (u))2 ≥ 0 and2(v) f (v)g(v)

1(v)− (g(v))2 ≥ 0,

hence ∫ x

a(x−u)−1EEE (w(x−u) ; p)( f (u))2du

≤∫ x

a(x−u)−1EEE (w(x−u) ; p)

2(u)1(u)

f (u)g(u)du,


and ∫ x

a(x− v)−1EEE (w(x− v) ; p)(g(v))2dv

≤∫ x

a(x− v)−1EEE (w(x− v) ; p)

2(v)1(v)

f (v)g(v)dv,

which imply ( f 2)(x; p) ≤ ( (2 f g/1))(x; p),(g2)(x; p) ≤ (

(2 f g/1))(x; p).

Multiplying above inequalities we obtain (4.14). �

Corollary 4.3 Suppose that the assumptions of Corollary 4.1 hold with > 0. Then( f 2

)(x; p)

(g2

)(x; p)

( f g) (x; p)( f g

)(x; p)

≤ MNmn

.

Remark 4.2 As before, choosing particular values of parameters in Theorem 4.2 andTheorem 4.3, known Polya-Szego type inequalities for several fractional integral operatorscan be deduced, such as inequalities for the Riemann-Liouville fractional integral operatorin [114, Lemma 3.3, Lemma 3.4] if we set w = p = 0.

4.2 Chebyshev Type Fractional IntegralInequalities and an ExtendedGeneralized Mittag-Leffler Function

Using Polya-Szego type inequality in Theorem 4.1, we obtain following Chebyshev in-equalities based on [120] and [114].

A simplified notation EEE and f as in (4.7) and (4.8) is used here.

Theorem 4.4 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,g,1,2,1 and 2 be positive integrable functions on [0,) functions satisfying

0 < 1(u) ≤ f (u) ≤ 2(u), 0 < 1(u) ≤ g(u) ≤ 2(u), u ∈ [a,x].

Suppose also > 0. Then∣∣(1)(x; p)( f g

)(x; p)+

(1

)(x; p)( f g) (x; p)

−( f ) (x; p)(g

)(x; p)− (g)(x; p)

( f

)(x; p)

∣∣≤ ∣∣G , ( f ,1,2)(x; p)+G , ( f ,1,2)(x; p)

∣∣ 12

× ∣∣G , (g,1,2)(x; p)+G ,(g,1,2)(x; p)∣∣ 1

2 , (4.15)

4.2 CHEBYSHEV TYPE INEQUALITIES AND MITTAG-LEFFLER FUNCTION 43

where

G , (u,v,w)(x; p) =

(1

)(x; p) [( (v+w)u)(x; p)]2

4(vw) (x; p)−(u)(x; p)

(u

)(x; p). (4.16)

Proof. Let f and g be two positive integrable function on [0,). For u,v ∈ [a,t] we defineA(u,v) as

A(u,v) = ( f (u)− f (v))(g(u)−g(v)),that is

A(u,v) = f (u)g(u)+ f (v)g(v)− f (u)g(v)− f (v)g(u).Multiplying above equality by


and integrating, we obtain∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)A(u,v)dudv

=(1

)(x; p)( f g)(x; p)+ (1)(x; p)

( f g

)(x; p)

−( f ) (x; p)(g

)(x; p)− ( f

)(x; p)(g)(x; p).

(4.17)By the Cauchy-Schwartz inequality for double integrals we have∣∣∣∣

∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)A(u,v)dudv

∣∣∣∣≤[∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)( f (u))2dudv

+∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)( f (v))2dudv

−2∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p) f (u) f (v)dudv

] 12

×[∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)(g(u))2dudv

+∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)(g(v))2dudv

−2∫ x

a

∫ x

a(x−u)−1(x− v)−1EEE (w(x−u) ; p)EEE (w(x− v) ; p)g(u)g(v)dudv

] 12

=[(1

)(x; p)

( f 2)(x; p)+ (1)(x; p)

( f 2) (x; p)

−2( f ) (x; p)( f

)(x; p)

] 12

×[(1)(x; p)

(g2)(x; p)+ (1)(x; p)

(g2)(x; p)

−2(g)(x; p)(g

)(x; p)

] 12 (4.18)


For 1(t) = 2(t) = g(t) = 1 by Theorem 4.1 follows

( f 2) (x; p) ≤ [( (1 +2) f ) (x; p)]2

4(12) (x; p).

This implies(1

)(x; p)

( f 2) (x; p)− ( f ) (x; p)

( f

)(x; p)

≤(1

)(x; p) [( (1 +2) f ) (x; p)]2

4(12) (x; p)− ( f ) (x; p)

( f

)(x; p)

= G , ( f ,1,2)(x; p) (4.19)

and

(1)(x; p)( f 2) (x; p)− ( f ) (x; p)

( f

)(x; p)

≤ (1)(x; p)[( (1 +2) f

)(x; p)

]24(12

)(x; p)

− ( f ) (x; p)( f

)(x; p)

= G , ( f ,1,2)(x; p) (4.20)

Applying the same procedure for 1(t) = 2(t) = f (t) = 1, we get(1

)(x; p)

(g2

)(x; p)− (g)(x; p)

(g

)(x; p)

≤ G , (g,1,2)(x; p) (4.21)

(1)(x; p)(g2

)(x; p)− (g)(x; p)

(g

)(x; p)

≤ G , (g,1,2)(x; p) (4.22)

Finally, considering (4.17) to (4.22), we arrive at the desired result in (4.15). This com-pletes the proof. �

Setting = in Theorem 4.4, next inequality follows.

Corollary 4.4 Suppose that the assumptions of Theorem 4.4 hold. Then

|(1)(x; p)( f g) (x; p)− ( f ) (x; p)(g)(x; p)|

≤ |G , ( f ,1,2)(x; p)G , (g,1,2)(x; p)| 12 .

If we set 1 = m, 2 = M, 1 = n and 2 = N in the previous corollary, then we obtain

G , ( f ,m,M)(x; p) =(M−m)2

4mM[( f ) (x; p)]2 ,

G , (g,n,N)(x; p) =(N−n)2

4nN[(g)(x; p)]2 .


Corollary 4.5 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let fand g be two positive integrable functions on [0,) satisfying

0 < m ≤ f (u) ≤ M < , 0 < n ≤ g(u)≤ N < , u ∈ [a,x].

Then|(1)(x; p)( f g) (x; p)− ( f ) (x; p)(g)(x; p)|

≤ (M−m)(N−n)4√

mMnN( f ) (x; p)(g)(x; p).

Remark 4.3 Setting w = p = 0 (and a = 0) in Theorem 4.4, Corollary 4.4 and Corollary4.5 we get Polya-Szego type inequalities for the Riemann-Liouville fractional integral op-erator given in [114, Theorem 3.6, Theorem 3.7, Corollary 3.4].

Recently, in [111] Nikolova and Varosanec generalized results from [114] for any twolinear isotonic functionals. Here, we will give another approach. In the next theorem wewill use Karamata’s estimations of the Chebyshev quotient ([89]),

1K2 ≤

(1

b−a

∫ ba f (x)dx

)(1

b−a

∫ ba g(x)dx

)1

b−a

∫ ba f (x)g(x)dx

≤ K2, (4.23)

where

K =√

mn+√

MN√mN +

√Mn

, (4.24)

and the result (4.6) by Pecaric and Peric ([120]). In this way we will obtain even betterupper and lower estimations than those in Corollary 4.5.

Theorem 4.5 Suppose that the assumptions of Corollary 4.5 hold. Then

1K2 ≤ (1)(x; p)( f g) (x; p)

( f ) (x; p)(g)(x; p)≤ K2 (4.25)

where K is given by (4.24).

Proof. Without loss of generality we can assume

1 ≤ f (u) ≤ 1, 1 ≤ g(u) ≤ 2

for every u ∈ [a,t] and some 1,2 ≥ 1. From the obvious inequality

[1− f (u)] [ f (v)−1] [2g(v)−g(u)]≥ 0

we obtain

12 f (v)g(v)− 12g(v)− 1 f (v)g(u)+ 1g(u)−2 f (u) f (v)g(v)+ f (u) f (v)g(u)+ 2 f (u)g(v)− f (u)g(u) ≥ 0.

Multiplying above inequality by



and then integrating, we have

12 (1)(x; p)( f g) (x; p)− 12 (1)(x; p)(g)(x; p)−1 ( f ) (x; p)(g)(x; p)+ 1 (1)(x; p)(g)(x; p)−2 ( f ) (x; p)( f g) (x; p)+ ( f ) (x; p)( f g)(x; p)+2 ( f ) (x; p)(g)(x; p)− (1)(x; p)( f g) (x; p)) ≥ 0

from which follows

2 [1 (1)(x; p)− ( f ) (x; p)]+ 1 [( f ) (x; p)− (1)(x; p)]2 [1 (1)(x; p)− ( f ) (x; p)]+ ( f ) (x; p)− (1)(x; p)

≤ ( f g) (x; p)(g)(x; p)

. (4.26)

Similarly, from[1− f (u)] [ f (v)−1] [2g(u)−g(v)]≥ 0

follows

( f g)(x; p)(g)(x; p)

≤1 (1)(x; p)− ( f ) (x; p)+ 12 [( f ) (x; p)− (1)(x; p)]1 (1)(x; p)− ( f ) (x; p)+ 2 [( f ) (x; p)− (1)(x; p)]

. (4.27)

Hence, from (4.26) and (4.27) we have

2

[1− ( f )(x;p)

( 1)(x;p)

]+ 1

[( f )(x;p)( 1)(x;p) −1

]2

[1− ( f )(x;p)

(1)(x;p)

]+ ( f )(x;p)

( 1)(x;p) −1

≤ ( f g) (x; p)(g)(x; p)

≤1− ( f )(x;p)

( 1)(x;p) + 12

[( f )(x;p)( 1)(x;p) −1

]1 − ( f )(x;p)

(1)(x;p) + 2

[( f )(x;p)(1)(x;p) −1

] .

Next, we define functions h,H : [1,1] → (0,) by

H(t) =1t1− t + 12(t−1)1− t + 2(t −1)

, h(t) =1

H(1/t).

For t1 =√1(

√1+√2)√12+1 it is straightforward to check that

maxt∈[1,1]

H(t) = H(t1) =(

1+√12√1 +

√2

)2

= K2

andmin

t∈[1,1]h(t) = h(1/t1) = 1/H(t1) = 1/K2.


Using (4.28) we obtain

h

(( f ) (x; p)(1)(x; p)

)≤ (1)(x; p)( f g) (x; p)

( f ) (x; p)(g)(x; p)≤ H

(( f ) (x; p)(1)(x; p)

)from which follows (4.25). �

Corollary 4.6 If the assumptions of Theorem 4.5 hold, then

− (M−m)(N−n)(√mN +

√Mn

)2 (1)(x; p)( f g) (x; p)

≤ − (M−m)(N−n)(√mn+

√MN

)2 ( f ) (x; p)(g)(x; p)

≤ (1)(x; p)( f g) (x; p)− ( f ) (x; p)(g)(x; p)

≤ (M−m)(N−n)(√mn+

√MN

)2 (1)(x; p)( f g)(x; p)

≤ (M−m)(N−n)(√mN +

√Mn

)2 ( f ) (x; p)(g)(x; p). (4.28)

Proof. As in [120, Corrolary 1], we see that direct consequences of (4.25) are the first andthe last inequality in (4.28). From the lower bound in (4.25) we have[

1K2 −1

](1)(x; p)( f g) (x; p)

≤ (1)(x; p)( f g) (x; p)− ( f ) (x; p)(g)(x; p)

from which follows the second inequality in (4.28). Analogously, from the upper bound in(4.25) follows the third inequality in (4.28). �

Remark 4.4 If we observe results from Corollary 4.5 and Corollary 4.6 we can see thatthe upper estimate from Corollary 4.6 is better, i.e. inequality

(M−m)(N−n)(√mN +

√Mn

)2 ≤ (M−m)(N−n)4√

mMnN

is equivalent with inequality

0 ≤(√

Mn−√mN

)2.

The upper estimates are equal if and only if M/m = N/n.The lower estimate from Corollary 4.6 is also better, i.e. inequality

− (M−m)(N−n)4√

mMnN≤− (M−m)(N−n)(√

mn+√

MN)2

is equivalent with inequality

0 ≤(√

mn−√MN

)2.

The lower estimates are equal if and only if m = M and n = N.

Chapter5Minkowski Type FractionalIntegral InequalitiesAssociated with theMittag-Leffler Function

Motivated by the papers [29, 142], where authors have proved Minkowski type integralinequalities, we present generalized corresponding results using our extended generalizedMittag-Leffler function with its fractional integral operators. We will need the Minkowskiintegral inequality and its reverse versions:

Theorem 5.1 (THE MINKOWSKI INTEGRAL INEQUALITY) Let p ≥ 1 and let f ,g ∈Lp[a,b]. Then

(∫ b

a| f (x)+g(x)|pdx

) 1p

≤(∫ b

a| f (x)|pdx

) 1p

+(∫ b

a|g(x)|pdx

) 1p

. (5.1)

Theorem 5.2 [29] Let p ≥ 1 and let f ,g ∈ Lp[a,b] be positive functions satisfying

0 < m ≤ f (x)g(x) ≤ M for x ∈ [a,b]. Then

(∫ b

a( f (x))pdx

) 1p

+(∫ b

a(g(x))pdx

) 1p

≤ M(m+2)+1(m+1)(M +1)

(∫ b

a( f (x)+g(x))pdx

) 1p

.

49

50 5 MINKOWSKI TYPE FRACTIONAL INTEGRAL INEQUALITIES

Theorem 5.3 [142] Let p ≥ 1 and let f ,g ∈ Lp[a,b] be positive functions satisfying

0 < m ≤ f (x)g(x) ≤ M for x ∈ [a,b]. Then

(∫ b

a( f (x))pdx

) 2p

+(∫ b

a(g(x))pdx

) 2p

≥ M(m−1)+m+1M

(∫ b

a( f (x))pdx

) 1p(∫ b

a(g(x))pdx

) 1p

.

The right-sided versions of following inequalities can be established and proved anal-ogously by using the right-sided fractional integral operators w, ,c,q,r

b−, , , f and hw, ,c,q,rb−, , , f

defined with (2.13) and (2.24) respectively.

This chapter is based on our results from [16, 17].

5.1 Reverse Minkowski Type Inequalities Involvingan Extended Generalized Mittag-Leffler Function

We start this section by presenting the reverse fractional Minkowski integral inequality us-ing extended generalized Mittag-Leffler function E ,c,v,r

, , with the corresponding fractional

integral operator w, ,c,v,ra+, , , f (in real domain).


EEE(z;u) := E ,c,v,r , , (z;u) (5.2)

=

n=0

Bu( +nv,c− )B( ,c− )

(c)nv

(n+)zn

()nr,

( f )(x;u) :=(w, ,c,v,ra+, , , f

)(x;u) (5.3)

=∫ x

a(x− t)−1EEE(w(x− t) ;u) f (t)dt.

For proving our inequality, we follow methods as in the paper by L. Bougoffa ([29]),which we supplement with the necessary steps to generalize Theorem 5.2.

Theorem 5.4 Let w∈ R, r, , , > 0, c > > 0 with u≥ 0 and 0 < v≤ r+ . Let p≥ 1and let f ,g ∈ Lp[a,b] be positive functions satisfying

0 < m ≤ f (x)g(x)

≤ M, x ∈ [a,b]. (5.4)

5.1 REVERSE MINKOWSKI TYPE INEQUALITIES 51

Then[( f p)(x;u)]

1p +[(gp)(x;u)]

1p ≤ c1 [(( f +g)p)(x;u)]

1p , (5.5)

where

c1 =M(m+2)+1

(m+1)(M +1). (5.6)

Proof. From the hypothesis f (t)g(t) ≤ M we have

f (t) ≤ M[ f (t)+g(t)]−M f (t), t ∈ [a,b],

from which follows the inequality

(M +1)p( f (t))p ≤ Mp [ f (t)+g(t)]p , p ≥ 1, t ∈ [a,b].

Multiplying both sides of the above inequality by

(x− t)−1EEE(w(x− t) ;u)

and integrating on [a,x] with respect to the variable t, we obtain

(M +1)p( f p)(x;u) ≤ Mp(( f +g)p)(x;u),

from which we get

[( f p)(x;u)]1p ≤ M

M +1[(( f +g)p)(x;u)]

1p . (5.7)

Further, for the lower bound f (t)g(t) ≥ m we have

g(t) ≤ 1m

[ f (t)+g(t)]− 1m

g(t), t ∈ [a,b],

and (1+

1m

)p

(g(t))p ≤(

1m

)p

[ f (t)+g(t)]p , p ≥ 1, t ∈ [a,b].

Similarly, if we multiply above inequality by (x− t)−1EEE(w(x− t) ;u) and integrate on[a,x], then we get (

1+1m

)p

(gp)(x;u) ≤(

1m

)p

(( f +g)p)(x;u),

from which follows

[(gp)(x;u)]1p ≤ 1

m+1[(( f +g)p)(x;u)]

1p . (5.8)

The resulting inequality (5.5) now follows by adding (5.7) and (5.8). �

Next theorem is a consequence of the Minkowski integral inequality (as shown by Setet al. in [142]) and it is a generalization of Theorem 5.3.


Theorem 5.5 Suppose that assumptions of Theorem 5.4 hold. Then

[( f p)(x;u)]2p +[(gp)(x;u)]

2p ≥ c2 [( f p)(x;u)]

1p [(gp)(x;u)]

1p , (5.9)

where

c2 =M(m−1)+m+1

M. (5.10)

Proof. Taking the product of the inequalities (5.7) and (5.8) we obtain

(M +1)(m+1)M

[( f p)(x;u)]1p [(gp)(x;u)]

1p ≤ [(( f +g)p)(x;u)]

2p .

If we apply Minkowski’s inequality (5.1) on the right hand side, we get

(M +1)(m+1)M

[( f p)(x;u)]1p [(gp)(x;u)]

1p

≤[[( f p)(x;u)]

1p +[(gp)(x;u)]

1p

]2

= [( f p)(x;u)]2p +2 [( f p)(x;u)]

1p [(gp)(x;u)]

1p +[(gp)(x;u)]

2p .

From this we can easily obtain the inequality (5.9). �

5.2 Related Minkowski Type Inequalities

In the following, we continue with the generalizations of the reverse Minkowski type inte-gral inequalities. For more similar results related to the Minkowski inequality in fractionalcalculus operators point of view, we refer the readers to see [37, 90, 91, 128, 149, 145, 146].A simplified notation EEE and f as in (5.2) and (5.3) is used here.

We start with two theorems involving parameters p,q > 1 such that 1p + 1

q = 1.

Theorem 5.6 Suppose that assumptions of Theorem 5.4 hold. Let p,q > 1 with 1p + 1

q = 1.Then

[( f )(x;u)]1p [(g)(x;u)]

1q ≤

(Mm

) 1pq

(( f1p g

1q ))(x;u). (5.11)

Proof. From the hypothesis f (t)g(t) ≤ M we obtain

( f (t))1q ≤ M

1q (g(t))

1q ,

and after multiplication by ( f (t))1p we get

f (t) ≤ M1q ( f (t))

1p (g(t))

1q , t ∈ [a,b].

5.2 RELATED MINKOWSKI TYPE INEQUALITIES 53

Multiplying both sides of the above inequality by (x−t)−1EEE(w(x− t) ;u) and integratingon [a,x] with respect to the variable t, we derive

( f )(x;u) ≤ M1q (( f

1p g

1q ))(x;u),

and

[( f )(x;u)]1p ≤ M

1pq

[(( f

1p g

1q ))(x;u)

] 1p. (5.12)

Further, we obtain

g(t) ≤ m− 1p ( f (t))

1p (g(t))

1q

by using lower bound m≤ f (t)g(t) and multiplication by (g(t))

1q . Multiplying above inequality

by (x− t)−1EEE(w(x− t) ;u) and integrating on [a,x] we have

(g)(x;u) ≤ m− 1p (( f

1p g

1q ))(x;u)

from which we get

[(g)(x;u)]1q ≤ m− 1

pq

[(( f

1p g

1q ))(x;u)

] 1q. (5.13)

The inequality (5.11) now follows from the product of inequalities (5.12) and (5.13). �

In the next theorem we will need the well known Young’s inequality for products ofx,y ≥ 0 with p,q > 1 such that 1

p + 1q = 1:

xy ≤ xp

p+

yq

q, (5.14)

and following elementary inequality for x,y ≥ 0 and p > 1:

(x+ y)p ≤ 2p−1(xp + yp). (5.15)

Theorem 5.7 Suppose that assumptions of Theorem 5.4 hold. Let p,q > 1 with 1p + 1

q = 1.Then

(( f g))(x;u) ≤ 2p−1

p

(M

M +1

)p

(( f p +gp))(x;u)

+2q−1

q

(1

m+1

)q

(( f q +gq))(x;u). (5.16)

Proof. We follow the same steps as in the proof of Theorem 5.4 to obtain (5.7) and (5.8)from which follow

1p( f p)(x;u) ≤ Mp

p(M +1)p (( f +g)p)(x;u) (5.17)

and1q(gq)(x;u) ≤ 1

q(m+1)q (( f +g)q)(x;u). (5.18)


Using Young’s inequality (5.14) we have

f (t)g(t) ≤ ( f (t))p

p+

(g(t))q

q.

Multiplying both sides of the above inequality by (x−t)−1EEE(w(x− t) ;u) and integratingon [a,x] we get

(( f g))(x;u) ≤ 1p( f p)(x;u)+

1q(gq)(x;u). (5.19)

From (5.17), (5.18) and (5.19) follows

(( f g))(x;u) ≤ Mp

p(M +1)p (( f +g)p)(x;u)+1

q(m+1)q (( f +g)q)(x;u). (5.20)

Using elementary inequality (5.15) we obtain

(( f +g)p)(x;u) ≤ 2p−1(( f p +gp))(x;u) (5.21)

and(( f +g)q)(x;u) ≤ 2q−1(( f q +gq))(x;u). (5.22)

Hence, from (5.20), (5.21) and (5.22) follows (5.16). �

Theorem 5.8 Suppose that assumptions of Theorem 5.4 hold. Then

1M

(( f g))(x;u) ≤ 1(m+1)(M +1)

(( f +g)2)(x;u)

≤ 1m

(( f g))(x;u). (5.23)

Proof. From 0 < m ≤ f (t)g(t) ≤ M and 1

M ≤ g(t)f (t) ≤ 1

m we obatin

(m+1)g(t)≤ f (t)+g(t)≤ (M +1)g(t) (5.24)

and (M +1

M

)f (t) ≤ f (t)+g(t)≤

(m+1

m

)f (t). (5.25)

Multiplying inequalities (5.24) and (5.25) we get

1M

f (t)g(t) ≤(f (t)+g(t)

)2(M +1)(m+1)

≤ 1m

f (t)g(t).

Finally, multiplying above inequalities by (x− t)−1EEE(w(x− t) ;u) and integrating on[a,x] with respect to the variable t, we obtain inequalities (5.23). �

In the last theorem of this section, we will add a positive parameter and assume < m, i.e.

0 < < m ≤ f (x)g(x)

≤ M, x ∈ [a,b].

5.2 RELATED MINKOWSKI TYPE INEQUALITIES 55

Theorem 5.9 Suppose that assumptions of Theorem 5.4 hold. Let > 0 be such that < m in (5.4). Then

M +1M−

(( f − g)1p )(x;u) ≤ [( f p)(x;u)]

1p +[(gp)(x;u)]

1p

≤ m+1m−

(( f − g)1p )(x;u). (5.26)

Proof. From the given condition 0 < < m ≤ M, we have

m ≤ M ⇒ m+m≤ m +M ≤ M+M,

that ism−M ≤ M−m .

Now we havem−M+Mm− ≤ M−m+Mm−

andM +1M−

≤ m+1m−

.

Easily we get

m− ≤ f (t)− g(t)g(t)

≤ M−

from which we can obtain

[ f (t)− g(t)]p

(M− )p ≤ (g(t))p ≤ [ f (t)− g(t)]p

(m− )p .

Multiplying above inequalities by (x− t)−1EEE(w(x− t) ;u) and integrating on [a,x] withrespect to the variable t, we obtain

1(M− )p (( f − g)p)(x;u) ≤ (gp)(x;u)

≤ 1(m− )p (( f − g)p)(x;u),

followed by

1M−

[(( f − g)p)(x;u)]1p ≤ [(gp)(x;u)]

1p

≤ 1m−

[(( f − g)p)(x;u)]1p . (5.27)

Further, from 1M ≤ g(t)

f (t) ≤ 1m we get

m− m

≤ f (t)− g(t)f (t)

≤ M− M

,


from which we have

Mp [ f (t)− g(t)]p

(M− )p ≤ ( f (t))p ≤ mp [ f (t)− g(t)]p

(m− )p .

Again, multiplying above inequalities by (x−t)−1EEE(w(x− t) ;u) and integrating on [a,x]we obtain

Mp

(M− )p (( f − g)p)(x;u) ≤ ( f p)(x;u)

≤ mp

(m− )p (( f − g)p)(x;u),

followed by

MM−

[(( f − g)p)(x;u)]1p ≤ [( f p)(x;u)]

1p

≤ mm−

[(( f − g)p)(x;u)]1p . (5.28)

By adding the inequalities (5.27) and (5.28) follows the result (5.26). �

5.3 Further Generalizations of MinkowskiType Inequalities

In this section we give further generalizations of reverse Minkowski type integral inequali-ties for a generalized fractional integral operator h

w, ,c,v,ra+, , , f (2.23) containing an extended

Mittag-Leffler function E ,c,v,r , , in the kernel and then we prove some related fractional

Minkowski type integral inequalities.

For the reader’s convenience we will use a simplified notation

EEE(z;u) := E ,c,v,r , , (z;u),

(hhh f )(x;u) :=(

hw, ,c,v,ra+, , , f

)(x;u)

=∫ x

a(h(x)−h(t))−1E ,c,v,r

, , (w(h(x)−h(t)) ;u)h′(t) f (t)dt.

Theorem 5.10 Let w ∈ R, , ,,r > 0, c > > 0 with u ≥ 0 and 0 < v ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function and let f ,g,1,2 ∈ Lp[a,b]be positive functions satisfying

0 < 1(x) ≤ f (x)g(x)

≤ 2(x), x ∈ [a,b].

5.3 FURTHER GENERALIZATIONS OF MINKOWSKI TYPE INEQUALITIES 57

Then for p ≥ 1 the following inequality holds

[(hhh f p)(x;u)]1p +[(hhhgp)(x;u)]

1p

≤[(

hhh(

2

1+2

)p

( f +g)p)

(x;u)] 1

p

+[(

hhh(

11+1

)p

( f +g)p)

(x;u)] 1

p

. (5.29)

Proof. Let t ∈ [a,b]. From f (t)g(t) ≤ 2(t) we have

f (t) ≤ 2(t)[ f (t)+g(t)]−2(t) f (t),

and for p ≥ 1

( f (t))p ≤(

2(t)1+2(t)

)p

[ f (t)+g(t)]p . (5.30)


(h(x)−h(t))−1EEE (w(h(x)−h(t)) ;u)h′(t) (5.31)

and integrating on [a,x] with respect to the variable t, we obtain

(hhh f p) (x;u) ≤(

hhh((

2

1+2

)p

( f +g)p))

(x;u).

from which we get

[(hhh f p)(x;u)]1p ≤

[(hhh((

2

1+2

)p

( f +g)p))

(x;u)] 1

p

. (5.32)

Further, from f (t)g(t) ≥ 1(t) follows

g(t) ≤ 11(t)

[ f (t)+g(t)]− 11(t)

g(t),

and if p ≥ 1, then

(g(t))p ≤(

11+1(t)

)p

[ f (t)+g(t)]p . (5.33)

Similarly, if we multiply above inequality by (5.31) and integrate on [a,x], then we get

(hhhgp)(x;u) ≤(

hhh((

11+1

)p

( f +g)p))

(x;u)

and also

[(hhhgp) (x;u)]1p ≤

[(hhh((

11+1

)p

( f +g)p))

(x;u)] 1

p

. (5.34)

By adding (5.32) and (5.34), the resulting inequality (5.29) follows. �

Setting 1 and 2 to be constant functions, i.e. 1(x) = m and 2(x) = M for allx ∈ [a,b], we obtain the following result.


Corollary 5.1 Let w ∈ R, , ,,r > 0, c > > 0 with u ≥ 0 and 0 < v ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function and let f ,g ∈ Lp[a,b] bepositive functions satisfying the condition (5.4). Then for p ≥ 1 the following inequalityholds


1p

≤ c1 [(hhh( f +g)p) (x;u)]1p ,

where c1 is defined by (5.6).

Remark 5.1 If the function h is the identity function, then we obtain an inequality fromTheorem 5.4 for the generalized fractional operator w, ,c,v,r

a+, , , f .

Next theorem is a fractional generalization of Theorem 5.3. It follows by the use of theMinkowski integral inequality (5.1).

Theorem 5.11 Suppose the assumptions of Theorem 5.10 hold. Then

[(hhh f p) (x;u)]2p +[(hhhgp)(x;u)]

2p

≥[(

hhh(

1+2

2

)p

f p)

(x;u)] 1

p

[(hhh(1+1)p gp) (x;u)]

1p

−2 [(hhh f p)(x;u)]1p [(hhhgp)(x;u)]

1p . (5.35)

Proof. For p ≥ 1 and t ∈ [a,b], inequalities (5.30) and (5.33) can also be written as(1+2(t)2(t)

)p

(( f (t))p ≤ [ f (t)+g(t)]p

and

(1+1(t))p(g(t))p ≤ [ f (t)+g(t)]p .

If we multiply both sides of each inequality by (5.31), integrate on [a,x] with respect to thevariable t and use power to the 1

p , then we obtain

[(hhh((

1+2

2

)p

f p))

(x;u)] 1

p

≤ [(hhh( f +g)p) (x;u)]1p (5.36)

and

[(hhh((1+1)p gp))(x;u)]

1p ≤ [(hhh( f +g)p)(x;u)]

1p . (5.37)

Taking the product of the inequalities (5.36) and (5.37) we obtain

[(hhh((

1+2

2

)p

f p))

(x;u)] 1

p

[(hhh((1+1)p gp)) (x;u)]

1p

≤ [(hhh( f +g)p)(x;u)]2p .


If we apply Minkowski’s inequality on right hand side, we get

[(hhh((

1+2

2

)p

f p))

(x;u)] 1

p

[(hhh((1+1)p gp)) (x;u)]

1p

≤[((hhh f p) (x;u))

1p +((hhhgp) (x;u))

1p

]2.

From this we can easily obtain the inequality (5.35). �

If 1(x) = m and 2(x) = M for all x ∈ [a,b], then the next inequality follows.

Corollary 5.2 Suppose the assumptions of Corollary 5.1 hold. Then


2p

≥ c2 [(hhh f p)(x;u)]1p [(hhhgp) (x;u)]

1p ,

where c2 is defined by (5.10).

Remark 5.2 If the function h is the identity function, then we obtain Theorem 5.5, aninequality for the generalized fractional operator w, ,c,v,r

a+, , , f .

We continue with the generalizations of the reverse Minkowski type integral inequali-ties. Starting conditions that we will need in this section are those given in Corollary 5.1,where we have 0 < m ≤ f (x)

g(x) ≤ M.

Theorem 5.12 Suppose that assumptions of Corollary 5.1 hold. Let p,q > 1 with 1p + 1

q =1. Then

[(hhh f ) (x;u)]1p [(hhhg)(x;u)]

1q ≤

(Mm

) 1pq (

hhh(

f1p g

1q

))(x;u). (5.38)

Proof. Let t ∈ [a,b]. From f (t)g(t) ≤ M we obtain

( f (t))1q ≤ M

1q (g(t))

1q ,

and after multiplication by ( f (t))1p we get

f (t) ≤ M1q ( f (t))

1p (g(t))

1q .

If we multiply both sides of the above inequality by (5.31), integrate on [a,x] with respectto the variable t and use power to the 1

p , then we obtain

[(hhh f ) (x;u)]1p ≤ M

1pq

[(hhh(

f1p g

1q

))(x;u)

] 1p. (5.39)

Next from the lower bound m ≤ f (t)g(t) we have

g(t) ≤ m− 1p ( f (t))

1p (g(t))

1q .


Again, if we multiply both sides of the above inequality by (5.31), integrate on [a,x] withrespect to the variable t and use power to the 1

p , then we get

[(hhhg)(x;u)]1q ≤ m− 1

pq

[(hhh(

f1p g

1q

))(x;u)

] 1q. (5.40)

The inequality (5.38) now follows from the product of inequalities (5.39) and (5.40). �

In the next theorem we will need the inequality (5.15) along with Young’s inequality(5.14).

Theorem 5.13 Suppose that assumptions of Corollary 5.1 hold. Let p,q > 1 with 1p + 1

q =1. Then

(hhh( f g)) (x;u) ≤ 2p−1

p

(M

M +1

)p

(hhh( f p +gp))(x;u)

+2q−1

q

(1

m+1

)q

(hhh( f q +gq)) (x;u). (5.41)

Proof. Let t ∈ [a,b], p,q > 1 and 1p + 1

q = 1. From f (t)g(t) ≤ M we get

( f (t))p ≤(

MM +1

)p

[ f (t)+g(t)]p .

If we multiply both sides of the above inequality by (5.31) and integrate on [a,x] withrespect to the variable t, then we obtain

1p

(hhh f p)(x;u) ≤ 1p

(M

M +1

)p

(hhh( f +g)p)(x;u) (5.42)

Next from m ≤ f (t)g(t) we obtain

(g(t))q ≤(

1m+1

)q

[ f (t)+g(t)]q .

Again, if we multiply both sides of the above inequality by (5.31) and integrate on [a,x]with respect to the variable t, then we get

1q

(hhhgq)(x;u) ≤ 1q

(1

m+1

)q

(hhh( f +g)q)(x;u). (5.43)

Using Young’s inequality (5.14) we have

f (t)g(t) ≤ ( f (t))p

p+

(g(t))q

q.

Multiplying both sides of the above inequality by (5.31) and integrating on [a,x] we get

(hhh( f g)) (x;u) ≤ 1p

(hhh f p)(x;u)+1q

(hhhgq)(x;u). (5.44)


From (5.42), (5.43) and (5.44) we obtain

(hhh( f g)) (x;u) ≤ 1p

(M

M +1

)p

(hhh( f +g)p) (x;u)

+1q

(1

m+1

)q

(hhh( f +g)q)(x;u). (5.45)

Using elementary inequality (5.15) we obtain

(hhh( f +g)p)(x;u) ≤ 2p−1 (hhh( f p +gp)) (x;u) (5.46)

and

(hhh( f +g)q)(x;u) ≤ 2q−1 (hhh( f q +gq))(x;u). (5.47)

Hence, from (5.45), (5.46) and (5.47) we obtain (5.41). �

Next theorem needs a simple application of the given condition (5.4).

Theorem 5.14 Suppose that assumptions of Corollary 5.1 hold. Then

1M

(hhh( f g)) (x;u) ≤ 1(m+1)(M +1)

(hhh( f +g)2)(x;u)

≤ 1m

(hhh( f g)) (x;u). (5.48)

Proof. From 0 < m ≤ f (t)g(t) ≤ M and 1

M ≤ g(t)f (t) ≤ 1

m we obtain

(m+1)g(t)≤ f (t)+g(t) ≤ (M +1)g(t) (5.49)

and (M +1

M

)f (t) ≤ f (t)+g(t) ≤

(m+1

m

)f (t). (5.50)

Multiplying inequalities (5.49) and (5.50) we get

1M

f (t)g(t) ≤ ( f (t)+g(t))2

(M +1)(m+1)≤ 1

mf (t)g(t).

Inequalities (5.48) now follow if we multiply the above by (5.31) and integrate on [a,x]with respect to the variable t. �

Theorem 5.15 Suppose that assumptions of Corollary 5.1 hold. Let > 0 be such that < m in (5.4). Then

M +1M−

(hhh( f −g)

1p

)(x;u)

≤ [(hhh f p)(x;u)]1p +[(hhhgp) (x;u)]

1p

≤ m+1m−

(hhh( f −g)

1p

)(x;u). (5.51)


Proof. From the given condition 0 < < m ≤ M, we have

m−M ≤ M−m ,

from which follow

M +1M−

≤ m+1m−

,

m− ≤ f (t)−g(t)g(t)

≤ M−

and[ f (t)−g(t)]p

(M−)p ≤ (g(t))p ≤ [ f (t)−g(t)]p

(m−)p .

If we multiply above inequalities by (5.31), integrate on [a,x] with respect to the variable tand use power to the 1

p , then we get

1M−

[(hhh( f −g)p) (x;u)]1p

≤ [(hhhgp)(x;u)]1p

≤ 1m−

[(hhh( f −g)p)(x;u)]1p . (5.52)

Further, from 1M ≤ g(t)

f (t) ≤ 1m we get

m−m

≤ f (t)−g(t)f (t)

≤ M−M

,

from which we have

Mp [ f (t)−g(t)]p

(M−)p ≤ ( f (t))p ≤ mp [ f (t)−g(t)]p

(m−)p .

Again, multiplying above inequalities by (5.31), integrating on [a,x] and using power tothe 1

p , we obtain

MM−

[(hhh( f −g)p) (x;u)]1p

≤ [(hhh f p)(x;u)]1p

≤ mm−

[(hhh( f −g)p)(x;u)]1p . (5.53)

By adding the inequalities (5.52) and (5.53), we get (5.51). �

Remark 5.3 If in the obtained results we use the identity function for the function h, thenwe obtain Theorem 5.6 - Theorem 5.9 .

Chapter6Classical Integral Inequalitiesand the Mittag-LefflerFunction

This chapter is motivated with researches of classical integral inequalities by W. Liu et al.[99] and by Z. Dahmani [38], such as:

Theorem 6.1 [99, Theorem 4] Let f ,g be positive continuous functions on [a,b] suchthat f is decreasing and g is increasing. Then the following inequality∫ x

a( f (t)) dt∫ x

a( f (t)) dt

≥

∫ x

a(g(t)) ( f (t)) dt∫ x

a(g(t)) ( f (t)) dt

(6.1)

holds for every > 0 and ≥ > 0. If f is increasing, then (6.1) is reversed.

Theorem 6.2 [38, Theorem 3.6] Let ( fi)i=1,2,...,n and g be positive continuous functionson [a,b] such that ( fi)i=1,2,...,n are decreasing and g is increasing. Then the followinginequality

Ja+

[n

i�=s f ii f s (x)]Ja+

[g(x)n

i=1 f ii (x)]

Ja+

[g(x)n

i�=s f ii f s (x)]Ja+

[n

i=1 f ii (x)] ≥ 1 (6.2)

holds for every a < x ≤ b, > 0, > 0, ≥ s > 0, where s is a fixed integer in{1,2, . . . ,n}.

63

64 6 CLASSICAL INTEGRAL INEQUALITIES

The aim is to present corresponding results using our extended generalized Mittag-Leffler function with its fractional integral operators.

The right-sided versions of following inequalities can be established and proved anal-ogously by using the right-sided fractional integral operators w, ,c,q,r

b−, , , f and hw, ,c,q,rb−, , , f

defined with (2.13) and (2.24) respectively.

This chapter is based on our results from [18, 19].

6.1 Generalizations of Classical Integral InequalitiesInvolving an Extended Generalized Mittag-LefflerFunction

In this section we present certain classical integral inequalities using our extended gener-alized Mittag-Leffler function E ,c,q,r

, , with the corresponding fractional integral operator

w, ,c,q,ra+, , , f , in real domain.


EEE(z; p) := E ,c,q,r , , (z; p) (6.3)

=

n=0

Bp( +nq,c− )B( ,c− )

(c)nq

(n+)zn

()nr,

( f )(x; p) :=(w, ,c,q,ra+, , , f

)(x; p) (6.4)

=∫ x

a(x− t)−1EEE(w(x− t) ; p) f (t)dt.

For proving our inequalities, we follow similar methods as in the paper by W. Liu et al.([99]), which we supplement with the necessary steps to obtain generalized results.

Further extensions of these results are given in the following section.

Theorem 6.3 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ > 0 and x ∈ [a,b]. Let f ,g be positive continuous functions, monotonic in theopposite sense with f ∈ L [a,b] and g ∈ L [a,b]. Then the following inequality holds

( f )(x; p)( f )(x; p)

≥ ((g f ))(x; p)((g f ))(x; p)

. (6.5)

If f and g are monotonic functions in the same sense, then the inequality (6.5) is reversed.

6.1 GENERALIZATIONS OF CERTAIN INTEGRAL INEQUALITIES 65

Proof. Let f ,g be monotonic functions in the opposite sense, both positive and continuous.Then for u,v ∈ [a,x] we obtain

[(g(u)) − (g(v)) ][( f (v))− − ( f (u))−

]≥ 0, (6.6)

that is

(g(u))( f (v))− +(g(v))( f (u))−

≥ (g(u))( f (u))− +(g(v))( f (v))− .


(x− v)−1EEE((x− v) ; p)( f (v))

and integrating on [a,x] with respect to the variable v, we get

(g(u))( f )(x; p)+ ( f (u))−((g f ))(x; p)

≥ (g(u))( f (u))−( f )(x; p)+ ((g f ))(x; p).

Further multiplying by

(x−u)−1EEE((x−u) ; p)( f (u))

and then integrating on [a,x] with respect to the variable u, we have

((g f ))(x; p)( f )(x; p) ≥ ((g f ))(x; p)( f )(x; p)

from which follows (6.5).If f and g are monotonic in the same sense, then the reverse inequality of (6.5) can beproved analogously. �

For a special case of an increasing function on [a,b], g(x)= x−a, we have the followingcorollary.

Corollary 6.1 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ > 0 and x ∈ (a,b]. Let f ∈ L [a,b] be a positive continuous decreasingfunction. Then the following inequality holds

( f )(x; p)( f )(x; p)

≥ (((x−a) f ))(x; p)(((x−a) f ))(x; p)

. (6.7)

If f is increasing, then the inequality (6.7) is reversed.

Remark 6.1 If we consider special case of Mittag-Leffler function and its correspondinggeneralized fractional integral operator for p = w = 0, given in Remark 2.2, we obtain theleft-sided Riemann-Liouville fractional integral

Ja+ f (x) =1

()

∫ x

a(x− t)−1 f (t)dt.


If we also set = 1, then inequality (6.5) implies Theorem 6.1. By verifying the condition(6.6), it is easy to see that although Theorem 6.1 is stated only for the case of decreasingfunction f and increasing function g, inequality (6.1) remains valid even if f is increasingand g is decreasing function, hence monotone in the opposite sense.Similarly, for p = w = 0 and = 1, the inequality (6.7) implies [99, Theorem 3].

Theorem 6.4 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ > 0 and x ∈ [a,b]. Let f ,g be positive continuous functions, f ∈ L+ [a,b],g ∈ L [a,b], such that for u,v ∈ [a,x]

[(g(u))( f (v)) − (g(v))( f (u)) ][( f (v))− − ( f (u))−

]≥ 0. (6.8)


( f + )(x; p)( f + )(x; p)

≥ ((g f ))(x; p)((g f ))(x; p)

. (6.9)

If the condition (6.8) is reversed, then the inequality (6.9) is reversed.

Proof. According to condition (6.8) we arrive at

(g(u))( f (v))+− +(g(v))( f (u))+−

≥ (g(u))( f (v)) ( f (u))− +(g(v))( f (u)) ( f (v))− .

Multiplying the above by

(x− v)−1EEE((x− v) ; p)( f (v))


(g(u))( f + )(x; p)+ ( f (u))+−((g f ))(x; p)

≥ (g(u))( f (u))− ( f + )(x; p)+ ( f (u))((g f ))(x; p).

Once more, multiplying the above by

(x−u)−1EEE((x−u) ; p)( f (u))

and then integrating on [a,x] with respect to the variable u, we obtain

((g f ))(x; p)( f + )(x; p) ≥ ((g f ))(x; p)( f + )(x; p)

from which follows (6.9).If the condition (6.8) is reversed, then the reverse inequality of (6.9) can be proved analo-gously. �

Again, we have the following corollary for a special case g(x) = x−a.

6.1 GENERALIZATIONS OF CERTAIN INTEGRAL INEQUALITIES 67

Corollary 6.2 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ > 0 and x ∈ [a,b]. Let f ∈ L+ [a,b] be a positive continuous function suchthat for u,v ∈ [a,x]

[(u−a)( f (v)) − (v−a)( f (u)) ][( f (v))− − ( f (u))−

]≥ 0. (6.10)


( f + )(x; p)( f + )(x; p)

≥ (((x−a) f ))(x; p)(((x−a) f ))(x; p)

. (6.11)


Remark 6.2 For p = w = 0 and = 1, inequalities (6.9) and (6.11) imply [99, Theorem6] and [99, Theorem 5], respectively.

Next we present an essential integral inequality that we need in order to easily obtainTheorem 6.7.

Theorem 6.5 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letx ∈ [a,b]. Let f ,g,h ∈ L1[a,b] be positive continuous functions such that f/h and g aremonotonic in the opposite sense. Then the following inequality holds

( f )(x; p)(h)(x; p)

≥ (( f g))(x; p)((hg))(x; p)

. (6.12)

If f/h and g are monotonic in the same sense, then the inequality (6.12) is reversed.

Proof. From hypotheses on functions, for u,v ∈ [a,x] we have

[g(u)−g(v)][

f (v)h(v)

− f (u)h(u)

]≥ 0,

that is

g(u)f (v)h(v)

+g(v)f (u)h(u)

≥ g(u)f (u)h(u)

+g(v)f (v)h(v)

.


(x− v)−1EEE((x− v) ; p)h(v)


g(u)( f )(x; p)+f (u)h(u)

((gh))(x; p) ≥ g(u)f (u)h(u)

(h)(x; p)+ ((g f ))(x; p).

Again, multiplying the above by

(x−u)−1EEE((x−u) ; p)h(u)


and then integrating on [a,x] with respect to the variable u, we arrive at

((gh)(x; p)( f )(x; p) ≥ ((g f ))(x; p)(h)(x; p)

from which follows (6.12).If f/h and g are monotonic functions in the same sense, then the reverse inequality of(6.12) can be proved analogously. �

The counterpart of the previous result follows, where we assume f (x) ≤ h(x). Hence,inequality (6.12) remains satisfied if g is replaced by f −1.

Theorem 6.6 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let ≥ 1 and x ∈ [a,b]. Let f ,h ∈ L [a,b] be positive continuous functions such that f/hand f are monotonic in the opposite sense, with f (x) ≤ h(x) on [a,b]. Then the followinginequality holds

( f )(x; p)(h)(x; p)

≥ ( f )(x; p)(h)(x; p)

. (6.13)

If f/h and f are monotonic functions in the same sense, then the inequality (6.13) is re-versed.

Proof. Assume that f/h is a decreasing function and f an increasing one. Then for ≥ 1function f −1 is also increasing. By applying Theorem 6.5 we obtain

( f )(x; p)(h)(x; p)

≥ (( f ))(x; p)((h f −1))(x; p)

.

This together with the assumption f (x) ≤ h(x) lead to (6.13). Analogously we can provethe case when f/h is increasing and f decreasing, and obtain reversed inequality if f/hand f are monotonic in the same sense. �

In the last theorem of this section we involve a convex function in the inequality.

Theorem 6.7 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letx ∈ [a,b]. Let f ,g,h ∈ L1[a,b] be positive continuous functions such that f/h is decreasingfunction and f ,g are increasing, with f (x) ≤ h(x) on [a,b]. Let be a convex function on[0,] with (0) = 0. Then the following inequality holds

( f )(x; p)(h)(x; p)

≥ ((( f )g))(x; p)(((h)g))(x; p)

. (6.14)

Proof. The function (x)x is increasing since is a convex function on [0,] with (0) = 0.

From the assumption f (x) ≤ h(x) with the positivity of f and h, we get

( f (x))f (x)

≤ (h(x))h(x)

.

Further, since f ,g and (x)x are increasing then the following function

( f (x))f (x)

g(x)

6.2 EXTENSIONS OF CERTAIN INTEGRAL INEQUALITIES 69

is also increasing. Hence((( f )g))(x; p)(((h)g))(x; p)

=(( ( f )

f f g))(x; p)

(( (h)h hg))(x; p)

≤(( ( f )

f f g))(x; p)

(( ( f )f hg))(x; p)

,

and by applying Theorem 6.5 for f , h, ( f )f g we obtain

((( f )g))(x; p)(((h)g))(x; p)

≤ (( f ))(x; p)((h))(x; p)

.

�

Corollary 6.3 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letx ∈ [a,b]. Let f ,h ∈ L1[a,b] be positive continuous functions such that f/h is decreasingfunction and f is increasing, with f (x) ≤ h(x) on [a,b]. Let be a convex function on[0,] with (0) = 0. Then the following inequality holds

( f )(x; p)(h)(x; p)

≥ ((( f ))(x; p)(((h))(x; p)

. (6.15)

Remark 6.3 If we set p = w = 0 and = 1, then Theorem 6.5, Theorem 6.6, Theorem6.7 and Corollary 6.3 generalize Theorem 7, Theorem 8, Theorem 10 and Theorem 9 from[99], respectively.

6.2 Extensions of Classical Integral InequalitiesInvolving an Extended GeneralizedMittag-Leffler Function

We continue to further extend the previously presented integral inequalities. Again we useour extended generalized Mittag-Leffler function with the corresponding fractional inte-gral operator (in real domain) applied on ( fi)i=1,2,...,n. A simplified notation EEE and f asin (6.3) and (6.4) is used here.

First theorem is an extension of Theorem 6.3.

Theorem 6.8 Let w∈R, r, , , > 0, c > > 0 with p≥ 0 and 0< q≤ r+ . Let > 0, ≥ i > 0 for i = 1,2, . . . ,n and let x ∈ [a,b]. Let ( fi)i=1,2,...,n and g be positive continuousfunctions, such that ( fi)i=1,2,...,n are decreasing and g is increasing with ( fi)i=1,2,...,n ∈L [a,b] and g∈ L [a,b]. Then for the fixed integer s∈ {1,2, . . . ,n} the following inequalityholds (

(n

i�=s f ii f s))

(x; p)((n

i=1 f ii

))(x; p)

≥((g n

i�=s f ii f s))

(x; p)((g n

i=1 f ii

))(x; p)

. (6.16)

If ( fi)i=1,2,...,n are increasing and g is decreasing, then the inequality (6.16) also holds.If all functions are monotonic in the same sense, then the inequality (6.16) is reversed.


Proof. Let ( fi)i=1,2,...,n be decreasing and g increasing, all positive and continuous. Lets ∈ {1,2,3, ...,n}. Then for u,v ∈ [a,x] we obtain

[(g(u)) − (g(v)) ][( fs(v))−s − ( fs(u))−s

]≥ 0,

that is

(g(u))( fs(v))−s +(g(v))( fs(u))−s ≥ (g(u))( fs(u))−s +(g(v))( fs(v))−s .


(x− v)−1EEE((x− v) ; p)n

i=1

( fi(v))i


(g(u))(

(n

i�=s

f ii f s

))(x; p)+ ( fs(u))−s

(

(g

n

i=1

f ii

))(x; p)

≥ (g(u))( fs(u))−s(

(n

i=1

f ii

))(x; p)+

(

(g

n

i�=s

f ii f s

))(x; p).


(x−u)−1EEE((x−u) ; p)n

i=1

( fi(u))i

and then integrating on [a,x] with respect to the variable u, we have(

(g

n

i=1

f ii

))(x; p)

(

(n

i�=s

f ii f s

))(x; p)

≥(

(g

n

i�=s

f ii f s

))(x; p)

(

(n

i=1

f ii

))(x; p).

from which follows (6.16).Analogously we can prove the case when ( fi)i=1,2,...,n are increasing and g is decreasing,and obtain reversed inequality if all functions are monotonic in the same sense. �

Remark 6.4 For p = w = 0 we obtain the left-sided Riemann-Liouville fraction integralJa+ of order , as a special case of Mittag-Leffler function and its corresponding general-ized fractional integral operator. Therefore, Theorem 6.8 generalizes Theorem 6.2.On the other hand, if we set n = 1, then s = 1 which implies Theorem 6.3.

For g(x) = x− a, which is an increasing function on [a,b], we have the followingcorollary. In this case, the inequality (6.17) implies [38, Theorem 3.1].


Corollary 6.4 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ i > 0 for i = 1,2, . . . ,n and let x∈ (a,b]. Let ( fi)i=1,2,...,n be positive continuousdecreasing functions with ( fi)i=1,2,...,n ∈ L [a,b]. Then for the fixed integer s∈ {1,2, . . . ,n}the following inequality holds(

(n

i�=s f ii f s))

(x; p)((n

i=1 f ii

))(x; p)

≥(((x−a) n

i�=s f ii f s))

(x; p)(((x−a) n

i=1 f ii

))(x; p)

. (6.17)

If ( fi)i=1,2,...,n are increasing, then the inequality (6.17) is reversed.

Theorem 6.9 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ i > 0 for i = 1,2, . . . ,n and let x ∈ [a,b]. Let ( fi)i=1,2,...,n and g be positivecontinuous functions, ( fi)i=1,2,...,n ∈ L+ [a,b] and g ∈ L [a,b]. Let s ∈ {1,2, . . . ,n} befixed integer and for u,v ∈ [a,x]

[(g(u))( fs(v)) − (g(v))( fs(u)) ][( fs(v))−s − ( fs(u))−s

]≥ 0. (6.18)

Then the following inequality holds((n

i�=s f ii f +s

))(x; p)(

(n

i�=s f ii f +ss

))(x; p)

≥((g n

i�=s f ii f s))

(x; p)((g n

i=1 f ii

))(x; p)

. (6.19)


Proof. From the (6.18) we obtain

(g(u))( fs(v))+−s +(g(v))( fs(u))+−s

≥ (g(u))( fs(v)) ( fs(u))−s +(g(v))( fs(u))( fs(v))−s .


(x− v)−1EEE((x− v) ; p)n

i=1

( fi(v))i


(g(u))(

(n

i�=s

f ii f +s

))(x; p)+ ( fs(u))+−s

(

(g

n

i=1

f ii

))(x; p)

≥ (g(u))( fs(u))−s(

(n

i�=s

f ii f +ss

))(x; p)

+ ( fs(u))(

(g

n

i�=s

f ii f s

))(x; p).




i=1

( fi(u))i

and then integrating on [a,x] with respect to the variable u, we have

(

(g

n

i=1

f ii

))(x; p)

(

(n

i�=s

f ii f +s

))(x; p)

≥(

(g

n

i�=s

f ii f s

))(x; p)

(

(n

i�=s

f ii f +ss

))(x; p).

from which follows (6.19).If the condition (6.18) is reversed, then the reverse inequality of (6.19) can be provedanalogously. �

Remark 6.5 For p = w = 0, Theorem 6.9 generalizes [38, Theorem 3.10].Setting n = 1, Theorem 6.4 follows.

Corollary 6.5 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ i > 0 for i = 1,2, . . . ,n and let x ∈ (a,b]. Let ( fi)i=1,2,...,n ∈ L+ [a,b] bepositive continuous functions. Let s ∈ {1,2, . . . ,n} be fixed integer and for u,v ∈ [a,x]

[(u−a)( fs(v)) − (u− v)( fs(u)) ][( fs(v))−s − ( fs(u))−s

]≥ 0. (6.20)

Then the following inequality holds((n

i�=s f ii f +s

))(x; p)(

(n

i�=s f ii f +ss

))(x; p)

≥(((x−a) n

i�=s f ii f s))

(x; p)(((x−a) n

i=1 f ii

))(x; p)

. (6.21)


Theorem 6.10 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let > 0, ≥ i > 0 for i = 1,2, . . . ,n and let x ∈ [a,b]. Let f ,g,(hi)i=1,2,...,n ∈ L1[a,b] bepositive continuous functions such that f/hs and g are monotonic in the opposite sense,for s ∈ {1,2, . . . ,n}. Then the following inequality holds

((

f ni�=s hi

))(x; p)

( (ni=1 hi))(x; p)

≥((g f n

i�=s hi

))(x; p)

( (gni=1 hi))(x; p)

. (6.22)

If f/hs and g are monotonic in the same sense for s ∈ {1,2, . . . ,n}, the inequality (6.22) isreversed.



[g(u)−g(v)][

f (v)hs(v)

− f (u)hs(u)

]≥ 0,

that is

g(u)f (v)hs(v)

+g(v)f (u)hs(u)

≥ g(u)f (u)hs(u)

+g(v)f (v)hs(v)

.


(x− v)−1EEE((x− v) ; p)n

i=1

hi(v)


g(u)

(

(f

n

i�=s

hi

))(x; p)+

f (u)hs(u)

(

(g

n

i=1

hi

))(x; p)

≥ g(u)f (u)hs(u)

(

(n

i=1

hi

))(x; p)+

(

(g f

n

i�=s

hi

))(x; p).



i=1

hi(u)

and then integrating on [a,x] with respect to the variable u, we arrive at(

(g

n

i=1

hi

))(x; p)

(

(f

n

i�=s

hi

))(x; p)

≥(

(n

i=1

hi

))(x; p)

(

(g f

n

i�=s

hi

))(x; p).

from which follows (6.22).If f/hs and g are monotonic in the same sense for s ∈ {1,2, . . . ,n}, then the reverse in-equality of (6.22) can be proved analogously. �

Remark 6.6 For p = w = 0, Theorem 6.10 generalizes [38, Theorem 3.14].Setting n = 1, Theorem 6.5 follows.


6.3 Further Generalizations of Some ClassicalIntegral Inequalities

In this section we give further generalizations of several classical integral inequalities for ageneralized fractional integral operator h

w, ,c,q,ra+, , , f (2.23) containing an extended Mittag-

Leffler function E ,c,q,r , , in the kernel.

We use a simplified notation

EEE(z; p) := E ,c,q,r , , (z; p),

(hhh f )(x; p) :=(

hw, ,c,q,ra+, , , f

)(x; p)

=∫ x

a(h(x)−h(t))−1E ,c,q,r

, , (w(h(x)−h(t)) ; p)h′(t) f (t)dt.

Theorem 6.11 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ i > 0 fori = 1,2, . . . ,n and let x ∈ [a,b]. Let ( fi)i=1,2,...,n and g be positive continuous functions,such that ( fi)i=1,2,...,n are decreasing and g is increasing with ( fi)i=1,2,...,n ∈ L [a,b] andg ∈ L [a,b]. Then for the fixed integer s ∈ {1,2, . . . ,n} the following inequality holds

(hhh(n

i�=s f ii f s))

(x; p)(hhh(n

i=1 f ii

))(x; p)

≥(

hhh(g n

i�=s f ii f s))

(x; p)(hhh(g n

i=1 f ii

))(x; p)

. (6.23)

If ( fi)i=1,2,...,n are increasing and g is decreasing, then the inequality (6.23) also holds.If all functions are monotonic in the same sense, then the inequality (6.23) is reversed.

Proof. Let u,v ∈ [a,x]. Let ( fi)i=1,2,...,n be decreasing and g increasing, all positive andcontinuous. Let s ∈ {1,2,3, ...,n}. Then

[(g(u)) − (g(v)) ][( fs(v))−s − ( fs(u))−s

]≥ 0, (6.24)

hence

(g(u))( fs(v))−s +(g(v))( fs(u))−s ≥ (g(u))( fs(u))−s +(g(v))( fs(v))−s .


(h(x)−h(v))−1EEE (w(h(x)−h(v)) ; p)n

i=1

( fi(v))i h′(v) (6.25)

6.3 FURTHER GENERALIZATIONS 75


(g(u))(

hhh

(n

i�=s

f ii f s

))(x; p)+ ( fs(u))−s

(hhh

(g

n

i=1

f ii

))(x; p)

≥ (g(u))( fs(u))−s(

hhh

(n

i=1

f ii

))(x; p)+

(hhh

(g

n

i�=s

f ii f s

))(x; p).


(h(x)−h(u))−1EEE (w(h(x)−h(u)) ; p)n

i=1

( fi(u))i h′(u) (6.26)

and then integrating on [a,x] with respect to the variable u, we have(hhh

(g

n

i=1

f ii

))(x; p)

(hhh

(n

i�=s

f ii f s

))(x; p)

≥(

hhh

(g

n

i�=s

f ii f s

))(x; p)

(hhh

(n

i=1

f ii

))(x; p).

from which follows (6.23).Analogously we can prove the case when ( fi)i=1,2,...,n are increasing and g is decreasing,and obtain reversed inequality if all functions are monotonic in the same sense. �

Remark 6.7 If the function h is the identity function, then we obtain an inequality fromTheorem 6.8 for the generalized fractional operator w, ,c,v,r

a+, , , f .Also, if the h is the identity function and p = w = 0, then we obtain the left-sided

Riemann-Liouville fraction integral Ja+ of order , i.e. a special case of Mittag-Lefflerfunction and its corresponding generalized fractional integral operator. Therefore, Theo-rem 6.11 generalizes Theorem 6.2.

The conditions under which inequality (6.23) and reverse inequality hold are comple-mented by the remaining cases of monotonicity of functions in the theorem.

Next inequality follows by setting g(x) = x−a and it is a generalization of [38, Theo-rem 3.1].

Corollary 6.6 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ i > 0 fori = 1,2, . . . ,n and let x ∈ (a,b]. Let ( fi)i=1,2,...,n be positive continuous decreasing func-tions with ( fi)i=1,2,...,n ∈ L [a,b]. Then for the fixed integer s ∈ {1,2, . . . ,n} the followinginequality holds(

hhh(n

i�=s f ii f s))

(x; p)(hhh(n

i=1 f ii

))(x; p)

≥(

hhh((x−a) n

i�=s f ii f s))

(x; p)(hhh((x−a) n

i=1 f ii

))(x; p)

. (6.27)

If ( fi)i=1,2,...,n are increasing, then the inequality (6.27) is reversed.


For the next result we set n = 1 in Theorem 6.11.

Corollary 6.7 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ > 0 andx ∈ [a,b]. Let f ,g be positive continuous functions, monotonic in the opposite sense withf ∈ L [a,b] and g ∈ L [a,b]. Then the following inequality holds

(hhh f

)(x; p)

(hhh f ) (x; p)≥

(hhh(g f )

)(x; p)

(hhh(g f )) (x; p). (6.28)

If f and g are monotonic functions in the same sense, then the inequality (6.28) is reversed.

If additionally g(x) = x−a, then the following corollary holds.

Corollary 6.8 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ > 0 andx∈ (a,b]. Let f ∈ L [a,b] be a positive continuous decreasing function. Then the followinginequality holds

(hhh f

)(x; p)

(hhh f )(x; p)≥

(hhh((x−a) f )

)(x; p)

(hhh((x−a) f )) (x; p). (6.29)

If f is increasing, then the inequality (6.29) is reversed.

Remark 6.8 If the function h is the identity function, p = w = 0 and = 1, then inequal-ities (6.28) and (6.29) imply Theorem 6.1 and [99, Theorem 3], respectively.

Further, from the condition (6.24) with n = 1 (i.e. we have only one function f ) andfor u,v ∈ [a,x] we obtain

[(g(u)) − (g(v)) ][( f (v))− − ( f (u))−

]≥ 0.

Now it is easy to see that although Theorem 6.1 is stated only for the case of decreasingfunction f and increasing function g, inequality (6.1) remains valid even if f is increasingand g is decreasing function g, hence monotone functions in the opposite sense.

Theorem 6.12 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ i > 0 fori = 1,2, . . . ,n and let x ∈ [a,b]. Let ( fi)i=1,2,...,n and g be positive continuous functions,( fi)i=1,2,...,n ∈ L+ [a,b] and g ∈ L [a,b]. Let s ∈ {1,2, . . . ,n} be fixed integer and foru,v ∈ [a,x] let the condition (6.18) holds. Then(

hhh(n

i�=s f ii f +s

))(x; p)(

hhh(n

i�=s f ii f +ss

))(x; p)

≥(

hhh(g n

i�=s f ii f s))

(x; p)(hhh(g n

i=1 f ii

))(x; p)

. (6.30)



Proof. From the (6.18) we obtain

(g(u))( fs(v))+−s +(g(v))( fs(u))+−s

≥ (g(u))( fs(v)) ( fs(u))−s +(g(v))( fs(u))( fs(v))−s .

Multiplying both sides of the above inequality by (6.25) and integrating on [a,x] withrespect to the variable v, we get

(g(u))(

hhh

(n

i�=s

f ii f +s

))(x; p)+ ( fs(u))+−s

(hhh

(g

n

i=1

f ii

))(x; p)

≥ (g(u))( fs(u))−s(

hhh

(n

i�=s

f ii f +ss

))(x; p)+ (fs(u))

(hhh

(g

n

i�=s

f ii f s

))(x; p).

Further multiplying the above by (6.26) and then integrating on [a,x] with respect to thevariable u, we obtain(

hhh

(g

n

i=1

f ii

))(x; p)

(hhh

(n

i�=s

f ii f +s

))(x; p)

≥(

hhh

(g

n

i�=s

f ii f s

))(x; p)

(hhh

(n

i�=s

f ii f +ss

))(x; p).

from which follows (6.30).If the condition (6.18) is reversed, then the reverse inequality of (6.30) can be provedanalogously. �

Remark 6.9 If the function h is the identity function and p = w = 0, then Theorem 6.12generalizes [38, Theorem 3.10].

Corollary 6.9 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ i > 0 fori = 1,2, . . . ,n and let x ∈ (a,b]. Let ( fi)i=1,2,...,n ∈ L+ [a,b] be positive continuous func-tions. Let s∈ {1,2, . . . ,n} be fixed integer and for u,v∈ [a,x] let the condition (6.20) holds.Then (

hhh(n

i�=s f ii f +s

))(x; p)(

hhh(n

i�=s f ii f +ss

))(x; p)

≥(

hhh((x−a) n

i�=s f ii f s))

(x; p)(hhh((x−a) n

i=1 f ii

))(x; p)

. (6.31)


If we set n = 1 in Theorem 6.12, then we obtain the following inequality.

Corollary 6.10 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ > 0 andx ∈ [a,b]. Let f ,g be positive continuous functions, f ∈ L+ [a,b], g ∈ L [a,b], such thatfor u,v ∈ [a,x] the condition (6.8) holds. Then


(hhh f + )(x; p)(hhh f + )(x; p)

≥(hhh(g f )

)(x; p)

(hhh(g f )) (x; p). (6.32)


Corollary 6.11 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ > 0 andx ∈ (a,b]. Let f ∈ L+ [a,b] be positive continuous function such that for u,v ∈ [a,x] thecondition (6.10) holds. Then

(hhh f + )(x; p)(hhh f +) (x; p)

≥(hhh((x−a) f )

)(x; p)

(hhh((x−a) f )) (x; p). (6.33)


Remark 6.10 If the function h is the identity function, p = w = 0 and = 1, then in-equalities (6.32) and (6.33) imply [99, Theorem 6] and [99, Theorem 5], respectively.

Theorem 6.13 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let > 0, ≥ i > 0 fori = 1,2, . . . ,n and let x ∈ [a,b]. Let f ,g,(i)i=1,2,...,n ∈ L1[a,b] be positive continuousfunctions such that f/s and g are monotonic in the opposite sense, for s ∈ {1,2, . . . ,n}.Then the following inequality holds(

hhh(

f ni�=si

))(x; p)

(hhh(ni=1i))(x; p)

≥(

hhh(g f n

i�=si

))(x; p)

(hhh(gni=1i)) (x; p)

. (6.34)

If f/s and g are monotonic in the same sense for s ∈ {1,2, . . . ,n}, then the inequality(6.34) is reversed.


[g(u)−g(v)][

f (v)s(v)

− f (u)s(u)

]≥ 0,

that is

g(u)f (v)s(v)

+g(v)f (u)s(u)

≥ g(u)f (u)s(u)

+g(v)f (v)s(v)

.


(h(x)−h(v))−1EEE(w(h(x)−h(v)) ; p)n

i=1

i(v)h′(v)


g(u)

(hhh

(f

n

i�=s

i

))(x; p)+

f (u)s(u)

(hhh

(g

n

i=1

i

))(x; p)

≥ g(u)f (u)s(u)

(hhh

(n

i=1

i

))(x; p)+

(hhh

(g f

n

i�=s

i

))(x; p).



(h(x)−h(u))−1EEE(w(h(x)−h(u)) ; p)n

i=1

i(u)h′(u)

and then integrating on [a,x] with respect to the variable u, we arrive at(hhh

(g

n

i=1

i

))(x; p)

(hhh

(f

n

i�=s

i

))(x; p)

≥(

hhh

(n

i=1

i

))(x; p)

(hhh

(g f

n

i�=s

i

))(x; p).

from which follows (6.34).If f/s and g are monotonic in the same sense for s ∈ {1,2, . . . ,n}, then the reverse in-equality of (6.34) can be proved analogously. �

Remark 6.11 If the function h is the identity function and p = w = 0, then Theorem 6.13generalizes [38, Theorem 3.14].

If we set n = 1 in Theorem 6.13, then we obtain the following generalization of [99,Theorem 7].

Corollary 6.12 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function. Let x ∈ [a,b]. Let f ,g, ∈L1[a,b] be positive continuous functions such that f/ and g are monotonic in the oppositesense. Then the following inequality holds

(hhh f ) (x; p)(hhh) (x; p)

≥ (hhh( f g)) (x; p)(hhh(g))(x; p)

. (6.35)

If f/ and g are monotonic in the same sense, then the inequality (6.35) is reversed.

Next is a counterpart of the previous result, where we assume f (x) ≤ (x). Hence,inequality (6.35) remains satisfied if g is replaced by f −1.

Theorem 6.14 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b]→ R be a differentiable, strictly increasing function. Let ≥ 1 and x ∈ [a,b]. Letf , ∈ L [a,b] be positive continuous functions such that f/ and f are monotonic in theopposite sense, with f (x) ≤ (x) on [a,b]. Then the following inequality holds

(hhh f ) (x; p)(hhh) (x; p)

≥ (hhh f )(x; p)(hhh)(x; p)

. (6.36)

If f/ and f are monotonic functions in the same sense, then the inequality (6.36) isreversed.

Proof. Assume that f/ is a decreasing function and f an increasing one. Then for ≥ 1function f −1 is also increasing. By applying Corollary 6.12 we obtain

(hhh f ) (x; p)(hhh)(x; p)

≥ (hhh( f )) (x; p)(hhh( f −1))(x; p)

.


This together with the assumption f (x) ≤ (x) lead to (6.36). Analogously we can provethe case when f/ is increasing and f decreasing, and obtain reversed inequality if f/and f are monotonic in the same sense. �

For the last theorem we involve a convex function in the inequality.

Theorem 6.15 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Leth : [a,b] → R be a differentiable, strictly increasing function and x ∈ [a,b]. Let f ,g, ∈L1[a,b] be positive continuous functions such that f/ is decreasing function and f ,g areincreasing, with f (x) ≤ (x) on [a,b]. Let be a convex function on [0,] with (0) = 0.Then the following inequality holds


≥ (hhh(( f )g)) (x; p)(hhh(()g)) (x; p)

. (6.37)

Proof. The function (x)x is increasing since is a convex function on [0,] with (0) = 0.

From the assumption f (x) ≤ (x) with the positivity of f and , we get

( f (x))f (x)

≤ ((x))(x)

.

Further, since f ,g and (x)x are increasing then the following function

( f (x))f (x)

g(x)

is also increasing. Hence

(hhh(( f )g)) (x; p)(hhh(()g)) (x; p)

=

(hhh( ( f )

f f g))

(x; p)(hhh( ()

g))

(x; p)≤(

hhh( ( f )f f g)

)(x; p)(

hhh( ( f )f g)

)(x; p)

,

and by applying Corollary 6.12 for f , , ( f )f g we obtain

(hhh(( f )g)) (x; p)(hhh(()g)) (x; p)

≤ (hhh( f )) (x; p)(hhh()) (x; p)

.

�

Corollary 6.13 Let w ∈ R, , ,,r > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r+ . Let h :[a,b]→ R be a differentiable, strictly increasing function and x ∈ [a,b]. Let f , ∈ L1[a,b]be positive continuous functions such that f/ is decreasing function and f is increasing,with f (x) ≤ (x) on [a,b]. Let be a convex function on [0,] with (0) = 0. Then thefollowing inequality holds


≥ (hhh(( f ))) (x; p)(hhh(())) (x; p)

. (6.38)

Remark 6.12 If the function h is the identity function, p = w = 0 and = 1, then Theo-rem 6.14, Theorem 6.15 and Corollary 6.13 generalize Theorem 8, Theorem 10 and Theo-rem 9 from [99], respectively.

Chapter7Hadamard andFejer-Hadamard TypesFractional Integral InequalitiesAssociated with theMittag-Leffler Function

In this chapter Hadamard and Fejer-Hadamard inequalities for convex, relative convex, m-convex, (h−m)-convex, harmonically convex and harmonically (,h−m)-convex func-tions via fractional integrals involving Mittag-Leffler function are given.

This chapter is based on our results from [50, 58, 64, 83, 86, 87].

7.1 Hadamard and Fejer-Hadamard Inequalitiesfor Convex Functions

For a convex function f : I → R where I is an interval in R, the following inequality holds

f

(a+b

2

)≤ 1

b−a

∫ b

af (x)dx ≤ f (a)+ f (b)

2(7.1)

where a,b ∈ I and a < b.Inequality (7.1) is well known in literature as Hadamard inequality.

81

82 7 HADAMARD AND FEJER FRACTIONAL INTEGRAL INEQUALITIES

Fejer gave generalization of the Hadamard inequality known as Fejer-Hadamard inequalitystated as follows [67].For a convex function f : I → R where I is an interval in R, the following inequality holds

f

(a+b

2

)∫ b

ag(x)dx ≤ 1

b−a

∫ b

af (x)g(x)dx ≤ f (a)+ f (b)

2

∫ b

ag(x)dx, (7.2)

where g is a function which is integrable, nonnegative and symmetric about a+b2 .

For Riemann-Liouville fractional integrals the Hadamard inequality is given in next results.

Theorem 7.1 [140] Let f : [a,b]→ R be a function with 0 ≤ a < b and f ∈ L1[a,b]. If fis a convex function on [a,b], then the following inequality holds

f

(a+b

2

)≤ ( +1)

2(b−a)[Ja+ f (b)+ Jb− f (a)

]≤ f (a)+ f (b)2

(7.3)

with > 0.

Another version of the Hadamard inequality is given as follows.

Theorem 7.2 [141] Let f : [a,b]→ R be a function with 0 ≤ a < b and f ∈ L1[a,b]. If fis a convex function on [a,b], then the following inequality holds

f

(b+a

2

)≤ 2+1( +1)

(b−a)[J

( a+b2 )+ f (b)+ J

( a+b2 )−

f (a)] ≤ f (a)+ f (b)2

with > 0.

The generalized Hadamard inequality containing Mittag-Leffler function is given in thetheorem. The following notations will be used frequently

Hw′a+, (x; p) := (w′, ,c,q,r

a+, , , 1)(x; p), Hw′b−, (x; p) := (w′, ,c,q,r

b−, , , 1)(x; p),

while notations of fractional integral operators will be followed as it is.

Theorem 7.3 Let w ∈ R, r, , , > 0, > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R be a positive function with 0 ≤ a < b and f ∈ L1[a,b]. If f is a convexfunction on [a,b], then the following inequality for extended generalized fractional inte-gral holds

f

(a+b

2

)Hw′

a+, (b; p) (7.4)

≤(w′, ,c,q,r

a+, , , f )(b; p)+ (w′, ,c,q,rb−, , , f )(a; p)

2

≤(

f (a)+ f (b)2

)Hw′

b−, (a; p),

where w′ = w(b−a) .

7.1 INEQUALITIES FOR CONVEX FUNCTIONS 83

Proof. Since f is convex function on [a,b], for t ∈ [0,1] we have

f

((ta+(1− t)b)+ ((1− t)a+ tb)

2

)≤ f (ta+(1− t)b)+ f ((1− t)a+ tb)

2. (7.5)

Multiplying both sides of above inequality with t−1E ,c,q,r , , (wt ; p) we get

2t−1E ,c,q,r , , (wt ; p) f

(a+b

2

)

≤ t−1E ,c,q,r , , (wt ; p)( f (ta+(1− t)b)+ f ((1− t)a+ tb)).

Integrating with respect to t over [0,1] we have

2 f

(a+b

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+(1− t)b)dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ((1− t)a+ tb)dt.

If we put u = at +(1− t)b, then t = b−ub−a and if v = (1− t)a+ tb, then t = v−a

b−a . Thereforeby using Definition 2.2 one can have

f

(a+b

2

)Hw′

a+, (b; p) ≤(w′ , ,q,r

a+, , , f )(b; p)+ (w′, ,q,rb−, , , f )(a; p)

2. (7.6)

Again by using that f is convex function on [a,b], and for t ∈ [0,1] we have

f (ta+(1− t)b)+ f ((1− t)a+ tb)≤ t f (a)+ (1− t) f (b)+ (1− t) f (a)+ t f (b). (7.7)

Now multiplying with t−1E ,c,q,r , , (wt ; p) and integrating over [0,1] we get

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+(1− t)b)dt+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ((1− t)a+ tb)dt

≤ [ f (a)+ f (b)]∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt

from which by using change of variables as for (7.6), we get

(w′, ,q,ra+, , , f )(b; p)+ (w′, ,q,r

b−, , , f )(a; p) ≤ ( f (a)+ f (b))Hw′b−, (a; p). (7.8)

From the inequalities (7.6) and (7.8), the inequality (7.4) is obtained. �


Remark 7.1 In Theorem 7.3

(i) if p = 0, then we get [51, Theorem 2.1].

(ii) if w = p = 0, then we get [140, Theorem 2].

(iii) if w = p = 0 and = 1, then we get (7.1).

In the following we give the Fejer-Hadamard inequality for fractional integral operatorcontaining the extended generalized Mittag-Leffler function.

Theorem 7.4 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b]→R be a convex function with 0≤ a < b and f ∈ L1[a,b]. Also, let g : [a,b]→R bea function which is nonnegative, integrable and symmetric about a+b

2 . Then the followinginequality for the extended generalized fractional integral holds

f

(a+b

2

)(w′, ,c,q,r

a+, , , g)(b; p) (7.9)

≤(w′ , ,c,q,r

a+, , , f g)(b; p)+ (w′, ,c,q,rb−, , , f g)(a; p)

2

≤ f (a)+ f (b)2

(w′, ,c,q,rb−, , , g)(a; p),


Proof. Multiplying (7.5) with t−1E ,c,q,r , , (wt ; p)g(tb+(1− t)a) we get

2t−1E ,c,q,r , , (wt ; p) f

(a+b

2

)g(tb+(1− t)a)

≤ t−1E ,c,q,r , , (wt ; p)( f (ta+(1− t)b)+ f ((1− t)a+ tb))g(tb+(1− t)a).


2 f

(a+b

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)g(tb+(1− t)a)dt

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+(1− t)b)g(tb+(1− t)a)dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ((1− t)a+ tb)g(tb+(1− t)a)dt.

If we put u = at +(1− t)b, then t = b−ub−a and if v = (1− t)a+ tb, then t = v−a

b−a . Thereforeone can have

2 f

(a+b

2

)∫ b

a(b−u)−1 E ,c,q,r

, ,

(w

(b−ub−a

): p

)g(a+b−u)du

≤∫ b

a(b−u)−1 E ,c,q,r

, ,

(w

(b−ub−a

); p

)f (u)g(a+b−u)du

+∫ a

b(v−a)−1 E ,c,q,r

, ,

(w

(v−ab−a

); p

)f (v)g(a+b− v)dv.


From symmetry of function g about a+b2 we have g(a+b− x) = g(x), x ∈ [a,b], therefore

using this fact we have

f

(a+b

2

)(w′, ,q,r

a+, , ,g)(b; p) ≤(w′ , ,q,r

a+, , , f g)(b; p)+ (w′, ,q,rb−, , , f g)(a; p)

2. (7.10)

Now multiplying (7.7) with t−1E ,c,q,r , , (wt ; p)g(ta +(1− t)b) and integrating with re-

spect to t over [0,1] we get

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+(1− t)b)g(ta+(1− t)b)dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ((1− t)a+ tb)g(ta+(1− t)b)dt

≤ ( f (a)+ f (b))∫ 1

0t−1E ,c,q,r

, , (wt ; p)g(ta+(1− t)b)dt.

By change of variables as for (7.10), we get

(w′ , ,q,ra+ , , , f g)(b; p)+ (w′, ,q,r

b−, , , f g)(a; p) ≤ ( f (a)+ f (b)) (w′, ,q,rb−, , ,g)(a; p). (7.11)

From inequalities (7.11) and (7.10), we get inequality in (7.9). �


(i) if g = 1, then we get Theorem 7.3.

(ii) if p = 0, then we get [51, Theorem 2.2].

(iii) if w = p = 0, then we get [77, Theorem 2.2].

Another generalized version of the Hadamard inequality is given in next result.

Theorem 7.5 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R be a function such that f ∈ L1[a,b] with a < b. If f is convex on [a,b], thenthe following inequalities for extended generalized fractional integral operator holds

f

(a+b

2

)Hw′

( a+b2 )+, (b; p) (7.12)

≤[(

w′, ,c,q,r

( a+b2 )+, , ,

f

)(b; p)+

(w′, ,c,q,r

( a+b2 )−, , ,

f

)(a; p)

]

≤ f (a)+ f (b)2

Hw′( a+b

2 )−, (a; p)

where w′ = 2w(b−a) .


Proof. Since f is convex function, for t ∈ [0,1] we have

2 f

(a+b

2

)≤ f

(2− t

2a+

t2b

)+ f

(t2a+

2− t2

b

). (7.13)

Also from convexity of f we have

f

(2− t

2a+

t2b

)+ f

(t2a+

2− t2

b

)(7.14)

≤ 2− t2

f (a)+t2

f (b)+t2

f (a)+2− t

2f (b)

= f (a)+ f (b).

Multiplying (7.13) by t−1E ,c,q,r , , (wt ; p) on both sides and then integrating on [0,1] we

have

2 f

(a+b

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt (7.15)

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(2− t

2a+

t2b

)dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(t2a+

2− t2

b

)dt.

Putting u = 2−t2 a+ t

2b and v = t2a+ 2−t

2 b in (7.15), we have

2 f

(a+b

2

)∫ b

a+b2

(b− v)−1E ,c,q,r , , (w′(b− v) ; p)dv (7.16)

≤∫ b

a+b2

(b− v)−1E ,c,q,r , , (w′(b− v) ; p) f (v)dv

+∫ a+b

2

a(u−a)−1E ,c,q,r

, , (w′(u−a) ; p) f (u)du

By simplifying we get the first inequality of (7.12).Now multiplying (7.14) by t−1E ,c,q

, ,(wt ) on both sides and then integrating over [0,1]we have

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(2− t

2a+

t2b

)dt (7.17)

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(t2a+

2− t2

b

)

≤ [ f (a)+ f (b)]∫ 1

0(t−1)E ,c,q,r

, , (wt ; p)dt.


Putting u = 2−t2 a+ t

2b and v = t2a+ 2−t

2 b in (7.17), we have

∫ a+b2


, , (w′(u−a) ; p) f (u)du (7.18)

+∫ b

a+b2

(b− v)−1E ,c,q,r , , (w′(b− v) ; p) f (v)dv

≤∫ a+b

2


, , (w′(u−a) ; p)du,

which gives the second inequality of (7.12). �


(i) If p = 0, then we get [4, Theorem 3.9].

(ii) If w′ = p = 0 in Theorem 7.5, we get [141, Theorem 4].

(iii) If w′ = p = 0 and = 1, then we get the classical Hadamard inequality.

In the next results generalized fractional integral operators by a monotone increasing func-tion are utilized.

Theorem 7.6 Let f ,g : [a,b] → R, 0 < a < b be such that f is positive, f ∈ L1[a,b] andconvex on [a,b] and g differentiable and strictly increasing. Then the following inequalitiesfor extended generalized fractional integral operator defined in (2.23) and (2.24) hold:

f

(g(a)+g(b)

2

)(gw′ , ,c,q,r

a+, , , 1)

(b; p)

≤ 12

[(gw′, ,c,q,r

a+, , , f ◦ g)

(b; p)+(

gw′, ,c,q,rb−, , , f ◦ g

)(a; p)

]≤ f (g(a))+ f (g(b))

2

(gw′, ,c,q,r

b−, , , 1)

(a; p); w′ =w

(g(b)−g(a)).

Proof. Considering the following identity

g(a)+g(b)2

=12

[tg(a)+ (1− t)g(b)]+12

[(1− t)g(a)+ tg(b)].

For convex function f we have

2 f

(g(a)+g(b)

2

)≤ f (tg(a)+ (1− t)g(b))+ f ((1− t)g(a)+ tg(b)). (7.19)


Further from (7.19) one can obtain the following inequality:

2 f

(g(a)+g(b)

2

)∫ 1

0t−1E ,r,k,c

,, (wt ; p)dt (7.20)

≤∫ 1

0t−1E ,r,k,c

,, (wt ; p) f (tg(a)+ (1− t)g(b))dt

+∫ 1

0t−1E ,r,k,c

,, (wt ; p) f ((1− t)g(a)+ tg(b))dt.

Setting tg(a)+ (1− t)g(b) = g(x) that is t = g(b)−g(x)g(b)−g(a) and (1− t)g(a)+ tg(b) = g(y) that

is t = g(y)−g(a)g(b)−g(a) in (7.20), we get the following inequality:

2 f

(g(a)+g(b)

2

)(gw′, ,c,q,r

a+, , , 1)

(b; p) (7.21)

≤[(

gw′, ,c,q,ra+, , , f ◦ g

)(b; p)+

(gw′ , ,c,q,r

b−, , , f ◦ g)

(a; p)].

Further, by using the convexity of f , one can obtain

f (tg(a)+ (1− t)g(b))+ f ((1− t)g(a)+ tg(b)) (7.22)

≤ t f (g(a))+ (1− t) f (g(b))+ (1− t) f (g(a))+ t f (g(b)).

This leads to the following integral inequality:

∫ 1

0t−1E ,r,k,c

,, (wt ; p) f (tg(a)+ (1− t)g(b))dt (7.23)

+∫ 1

0t−1E ,r,k,c

,, (wt ; p) f ((1− t)g(a)+ tg(b))dt

≤∫ 1

0t−1E ,r,k,c

,, (wt ; p)(t f (g(a))+ (1− t) f (g(b)))dt

+∫ 1

0t−1E ,r,k,c

,, (wt ; p)((1− t) f (g(a))+ t f (g(b)))dt.

Setting tg(a)+ (1− t)g(b) = g(x) that is t = g(b)−g(x)g(b)−g(a) and (1− t)g(a)+ tg(b) = g(y) that

is t = g(y)−g(a)g(b)−g(a) in (7.23), and after calculation we get

(gw′, ,c,q,r

a+, , , f ◦ g)

(b; p)+(

gw′ , ,c,q,rb−, , , f ◦ g

)(a; p) (7.24)

≤ f (g(a))+ f (g(b))2

(gw′, ,c,q,r

b−, , , 1)

(a; p).

Combining (7.21) and (7.24), we get the required result. �

The Fejer-Hadamard inequality is given in the following result.


Theorem 7.7 Let f ,g,h : [a,b] → R, 0 < a < b be such that f and h are positive, f andh ∈ L1[a,b] and f convex on [a,b], while g is a differentiable and strictly increasing. Iff (g(a)+g(b)−g(x)) = f (g(x)), then the following inequalities for extended generalizedfractional integral operator defined in (2.23) and (2.24) hold:

f

(g(a)+g(b)

2

)(gw′ , ,c,q,r

a+, , , h ◦ g)

(b; p)

≤ 12

[(gw′, ,c,q,r

a+, , , (h ◦ g)( f ◦ g))

(b; p)+(

gw′, ,c,q,rb−, , , (h ◦ g)( f ◦ g)

)(a; p)

]≤ f (g(a))+ f (g(b))

2

(gw′, ,c,q,r

b−, , , h ◦ g)

(a; p), w′ =w

(g(b)−g(a)).

Proof. Multiplying both sides of (7.19) by t−1h(tg(a)+ (1− t)g(b))E ,r,k,c ,, (wt ; p) and

integrating on [0,1], we get

2 f

(g(a)+g(b)

2

)∫ 1

0t−1E ,r,k,c

,, (wt ; p)h(tg(a)+ (1− t)g(b))dt (7.25)

≤∫ 1

0t−1E ,r,k,c

,, (wt ; p)h(tg(a)+ (1− t)g(b)) f (tg(a)+ (1− t)g(b))dt

+∫ 1

0t−1E ,r,k,c

,, (wt ; p)h(tg(a)+ (1− t)g(b)) f ((1− t)g(a)+ tg(b))dt.

Setting tg(a) + (1− t)g(b) = g(x) that is t = g(b)−g(x)g(b)−g(a) and (1− t)g(a) + tg(b)=g(a) +

g(b)−g(x) in (7.25), and using f (g(a)+g(b)−g(x)) = f (g(x)), the following inequalityis obtained:

2 f

(g(a)+g(b)

2

)(gw′, ,c,q,r

a+, , , h ◦ g)

(b; p) (7.26)

≤[(

gw′ , ,c,q,ra+, , , (h ◦ g)( f ◦ g)

)(b; p)+

(gw′, ,c,q,r

b−, , , (h ◦ g)( f ◦ g))

(a; p)].

Multiplying t−1h((tg(a)+ (1− t)g(b))E ,r,k,c ,, (wt ; p) on both sides of (7.22) and inte-

grating over [0,1], we have

∫ 1

0t−1E ,r,k,c

,, (wt ; p)h((tg(a)+ (1− t)g(b)) f (tg(a)+ (1− t)g(b))dt (7.27)

+∫ 1

0t−1E ,r,k,c

,, (wt ; p)h((tg(a)+ (1− t)g(b)) f ((1− t)g(a)+ tg(b))dt

≤∫ 1

0t−1E ,r,k,c

,, (wt ; p)h((tg(a)+ (1− t)g(b))(t f (g(a))+ (1− t) f (g(b)))dt

+∫ 1

0t−1E ,r,k,c

,, (wt ; p)h((tg(a)+ (1− t)g(b))((1− t) f (g(a))+ t f (g(b)))dt.


Setting tg(a)+ (1− t)g(b)= g(x) and using f (g(a)+g(b)−g(x)) = f (g(x)) we get(gw′ , ,c,q,r

a+, , , (h ◦ g)( f ◦ g))

(b; p)+(

gw′, ,c,q,rb−, , , (h ◦ g)( f ◦ g)

)(a; p) (7.28)

≤ f (g(a))+ f (g(b))2

(gw′ , ,c,q,r

b−, , , (h ◦ g))

(a; p).

Combining (7.26) and (7.28), we get the required result. �

Remark 7.4 The Hadamard and the Fejer Hadamard inequalities given in Theorems 2.1–2.5 are special cases of theorems of this section.

7.2 Hadamard and Fejer-Hadamard Inequalitiesfor Relative Convex Functions

The following Hadamard inequality for relative convex functions via Riemann-Liouvillefractional integral operator is given in [112].

Theorem 7.8 Let f be a positive relative convex function integrable on [a,g(b)]. Thenthe following inequality holds

f

(a+g(b)

2

)≤ ( +1)

2(g(b)−a)[Ja+ f g(b)+ Jb− f (a)] ≤ f (a)+ f (g(b))

2

with > 0.

The generalized Hadamard inequality containing Mittag-Leffler function is given in thefollowing theorem.

Theorem 7.9 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,g(b)]→R be a function such that f ∈ L1[a,g(b)] with a < b. If f is relative convex on[a,g(b)], then for extended generalized fractional integral operator the following inequalityholds

f

(a+g(b)

2

)Hw′

a+, (g(b); p) (7.29)

≤ 12

[(w′ , ,c,q,ra+, , , f

)(g(b); p)+

(w′ , ,c,q,rg(b)−, , , f

)(a; p)

]≤ f (a)+ f (g(b))

2Hw′

g(b)−, (a; p),

where w′ = w(g(b)−a) .

7.2 INEQUALITIES FOR RELATIVE CONVEX FUNCTIONS 91

Proof. Since f is relative convex function, for t ∈ [0,1] we have

2 f

(a+g(b)

2

)≤ f (ta+(1− t)g(b))+ f ((1− t)a+ tg(b)). (7.30)

Multiplying (7.36) by 2t−1E ,c,q,r , , (wt ; p) on both sides and then integrating over [0,1]

we have

2 f

(a+g(b)

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt (7.31)

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+(1− t)g(b))dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ((1− t)a+ tg(b))dt.

Putting x = ta+(1− t)g(b) and y = (1− t)a+ tg(b) in the above inequality we have

2 f

(a+g(b)

2

)∫ a

g(b)

(g(b)− xg(b)−a

)−1

E ,c,q,r , ,

(w

(g(b)− xg(b)−a

); p

)( −dxg(b)−a

)(7.32)

≤∫ a

g(b)

(g(b)− xg(b)−a

)−1

E ,c,q,r , ,

(w

(g(b)− xg(b)−a

); p

)f (x)

( −dxg(b)−a

)

+∫ g(b)

a

(y−a

g(b)−a

)−1

E ,c,q,r , ,

(w

(y−a

g(b)−a

); p

)f (y)

(dy

g(b)−a

).

From the above inequality we get

2 f

(a+g(b)

2

)Hw′

a+, (g(b); p) ≤(w′, ,c,q,ra+, , , f

)(g(b); p)+

(w′, ,c,q,rg(b)−, , , f

)(a; p). (7.33)

Again using the relative convexity of f we have

f (ta+(1−t)g(b))+ f ((1−t)a+tg(b))≤ t f (a)+(1−t) f (g(b))+(1−t) f (a)+t f (g(b)).(7.34)

Multiplying (7.34) by t−1E ,c,q,r , , (wt ; p) on both sides and then integrating over [0,1] we

have

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+(1− t)g(b))dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ((1− t)a+ tg(b))dt

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p)t f (a)+ (1− t) f (g(b))dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p)(1− t) f (a)+ t f (g(b))dt.


Putting x = ta+(1− t)g(b) and y = (1− t)a+ tg(b) in above we get(w′, ,c,q,ra+, , , f

)(g(b); p)+

(w′, ,c,q,rg(b)−, , , f

)(a; p) ≤ [ f (a)+ f (g(b))]Hw′

g(b)−, (a; p). (7.35)

From (7.35) and (7.33), the inequality (7.29) is obtained. �

Remark 7.5

(i) If p = 0 in above Theorem 7.9, then we get [1, Theorem 2.8].

(ii) If w = p = 0 and q = 1 in Theorem 7.9, then we get Theorem 7.8.

In the next theorem the generalization of previous result is proved.

Theorem 7.10 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [g(a),g(b)] → R be a function such that f ∈ L1[g(a),g(b)] with a < b. If f is rela-tive convex on [g(a),g(b)], then for extended generalized fractional integral operator thefollowing inequality holds

f

(g(a)+g(b)

2

)Hw′

g(a)+, (g(b); p)

≤ 12

[(w′, ,c,q,rg(a)+, , , f

)(g(b); p)+

(w′ , ,c,q,rg(b)−, , , f

)(g(a); p)

]≤ f (g(a))+ f (g(b))

2Hw′

g(b)−, (g(a); p),

where w′ = w(g(b)−g(a)) .

Proof. Since f is relative convex function on [g(a),g(b)], for t ∈ [0,1] we have

2 f

(g(a)+g(b)

2

)≤ f (tg(a)+ (1− t)g(b))+ f ((1− t)g(a)+ tg(b)). (7.36)

Remaining proof of is on same lines as the proof of above theorem. �

Remark 7.6

(i) If p = 0 in above Theorem 7.10, then we get [1, Theorem 2.10].

(ii) If w = p = 0 in above Theorem 7.10, then we get [66, Corollary 1].

7.3 INEQUALITIES FOR m-CONVEX FUNCTIONS 93

7.3 Hadamard and Fejer-Hadamard Inequalitiesfor m-convex Functions

In this section we give Hadamard and Fejer-Hadamard inequalities for m-convex functionsvia extended Mittag-Leffler function.

Theorem 7.11 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [0,)→R be a positive function. Let a,b∈ [0,) with 0≤ a < mb and f ∈ L1[a,mb]. Iff is m-convex function on [a,mb], then the following inequality for the extended generalizedfractional integral holds

f

(a+mb

2

)Hw′

a+, (mb; p) (7.37)

≤(w′ , ,c,q,r

a+ , , , f )(mb; p)+m+1(mw′, ,c,q,rb−, , , f )( a

m ; p)

2

≤ m+1

2

⎡⎣ f (a)−m2 f

(a

m2

)mb−a

Hmw′b−,+1

( am

; p)

+(

f (b)+mf( a

m2

))Hmw′

b−,

( am

; p)]

,

where w′ = w(mb−a) .

Proof. Since f is m-convex function on [a,mb], for t ∈ [0,1] we have

f

((ta+m(1− t)b)+m

((1− t) a

m + tb)

2

)(7.38)

≤ f (ta+m(1− t)b)+mf((1− t) a

m + tb)

2.

Multiplying with t−1E ,c,q,r , , (wt ; p) the both sides of above inequality we get

2t−1E ,c,q,r , , (wt ; p) f

(a+mb

2

)

≤ E ,c,q,r , , (wt ; p)

(f (ta+m(1− t)b)+mf

((1− t)

am

+ tb))

.


2 f

(a+mb

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+m(1− t)b)dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p) f((1− t)

am

+ tb)

dt.


If we take u = at + m(1− t)b, then t = mb−umb−a and if v = (1− t) a

m + tb, then t = v− am

b− am

.

Therefore one can have

f

(a+mb

2

)Hw′

a+, (mb; p) ≤(w′, ,c,q,r

a+, , , f )(mb; p)+m+1(mw′, ,c,q,rb− , , , f )

(am ; p

)2

. (7.39)

Again by using that f is m-convex function we have

f (ta+m(1− t)b)+mf((1− t)

am

+ tb)

(7.40)

≤ t f (a)+m(1− t) f (b)+m2(1− t) f( a

m2

)+mt f (b).

Now multiplying with t−1E ,c,q,r , , (wt ; p) and integrating with respect to t over [0,1] we

get

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+m(1− t)b)dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p) f((1− t)

am

+ tb)

dt

≤[f (a)−m2 f

( am2

)]∫ 1

0tE ,c,q,r

, , (wt ; p)dt

+m[f (b)+mf

( am2

)]∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt.

From which by using change of variables as did for (7.39), we get

(w′, ,c,q,ra+, , , f )(mb; p)+m+1(mw′, ,c,q,r

b− , , , f )( am ; p)

2(7.41)

≤ m+1

2

⎡⎣ f (a)−m2 f

(a

m2

)mb−a

Hmw′b−,+1

( am

; p)

+(

f (b)+mf( a

m2

))Hmw′

b−,

( am

; p)]

.

From inequalities (7.39) and (7.41), we get the inequality (7.37). �


(i) if p = 0, then we get [50, Theorem 3].

(ii) if w = p = 0, m = 1, then we get [140, Theorem 2].

(iii) if m = 1, then we get Theorem 7.3.


Theorem 7.12 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,mb]→ R be a function such that f ∈ L1[a,mb] with 0 ≤ a < mb. If f is m-convex on[a,mb], then the following inequality for extended generalized fractional integral operatorholds

2 f

(a+mb

2

)Hw′

( a+mb2 )+, (mb; p) (7.42)

≤(w′, ,c,q,r

( a+mb2 )+, , ,

f

)(mb; p)+m+1

(mw′, ,c,q,r

( a+mb2 )−, , ,

f

)( am

; p)

≤ 1mb−a

(f (a)−m2 f

( am2

))Hw′

( a+mb2 )+,+1

(mb; p)

+m+1(

f (b)+mf( a

m2

))Hw′m

( a+mb2 )−,

( am

; p)

where w′ = 2w(mb−a) .

Proof. Since f is m-convex, for t ∈ [0,1] we have

2 f

(a+mb

2

)≤ f

(t2a+

2− t2

mb

)+mf

(2− t2m

a+t2b

). (7.43)

Also from m-convexity of f , we have

f

(t2a+m

2− t2

b

)+mf

(2− t2m

a+t2b

)(7.44)

≤ t2

(f (a)−m2 f

( am2

))+m

(f (b)+mf

( am2

)).

Multiplying (7.43) by t−1E ,c,q,r , , (wt ; p) on both sides and then integrating on [0,1] we

have

2 f

(a+mb

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt (7.45)

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(t2a+

2− t2

mb

)dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(2− t2m

a+t2b

)dt.

Putting u = t2a+ 2−t

2 mb and v = 2−t2m a+ t

2b in (7.45), we have

2 f

(a+mb

2

)∫ mb

a+mb2

(mb−u)−1E ,c,q,r , , (w′(mb−u) ; p)du

≤∫ mb

a+mb2

(mb−u)−1E ,c,q,r , , (w′(mb−u) ; p) f (u)du

+m+1∫ a+mb

2m

am

(v− a

m

)−1E ,c,q,r , ,

(mw′(v− a

m) : p

)f (v)dv.


By using (2.20) and Definition 2.2 we get first inequality of (7.42).Now multiplying (7.44) by t−1E ,c,q,r

, , (wt ; p) on both sides and then integrating withrespect to t over [0,1], we have

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(t2a+m

2− t2

b

)dt (7.46)

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(2− t2m

a+t2b

)

≤ 12

(f (a)−m2 f (

am2 )

)∫ 1

0tE ,c,q,r

, , (wt ; p)dt

+m(

f (b)+mf (am2 )

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt.

Putting u = t2a+m 2−t

2 b and v = 2−t2m a+ t

2b in (7.46), we have

∫ mb

a+mb2

(mb−u)−1E ,c,q,r , , (w′(mb−u) ; p) f (u)du (7.47)

+∫ a+mb

2m

am

(v− a

m

)−1E ,c,q,r , , (mw′

(v− a

m

); p) f (v)dv

≤ 12

(f (a)−m2 f (

am2 )

)∫ mb

a+mb2

(mb−u)E ,c,q,r , ,

(w′(mb−u) ; p

)dt

+m+1(

f (b)+mf (am2 )

)∫ a+mb2m

am

(v− a

m

)−1E ,c,q,r , , (mw′

(v− a

m

); p)dt.

By using (2.18) and Definition 2.2 we get second inequality of (7.42). �

Remark 7.8 In Theorem 7.12.


(ii) If m = 1, then we get Theorem 7.5.

(iii) If w = p = 0, m = 1 and = 1, then we get the classical Hadamard inequality.

Theorem 7.13 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [0,) → R be a m-convex function, a,b ∈ [0,) with 0 ≤ a < mb and f ∈ L1[a,mb].Also, let g : [a,mb] → R be a function which is nonnegative and integrable on [a,mb]. Iff (a+mb−mx) = f (x), then the following inequality for extended generalized fractionalintegral holds

f

(a+mb

2

)(w′, ,c,q,r

b−, , , g)( a

m; p)

(7.48)

≤ (1+m)2

(w′, ,c,q,rb−, ,+1, f g)

( am

; p)


≤ 12

⎡⎣ f (a)−m2 f

(a

m2

)mb−a

(w′ , ,c,q,rb−, ,+1,g)

( am

; p)

+m(

f (b)+mf( a

m2

))(w′, ,c,q,r

b−, , , g)( a

m; p)]

where w′ = w(b− a

m ) .

Proof. Proof of this theorem is on same lines as the proof of Theorem 7.18. �

In the next results generalized fractional integral operators by a monotone increasingfunction are utilized.

Theorem 7.14 Let f ,g : [a,b] → R, 0 < a < b, Range(g) ⊂ [a,b], be such that f is pos-itive and f ∈ L1[a,b], g differentiable and strictly increasing. If f be m-convex m ∈ (0,1]and g(a)<mg(b), then the following inequalities for fractional operators (2.23) and (2.24)hold:

f

(g(a)+mg(b)

2

)(g

,r,k,c ,, ,w′,a+1

)(g−1(mg(b)); p)

≤(

g ,r,k,c ,, ,w′,a+( f ◦ g)

)(g−1(mg(b)); p)+m+1

(gmw′, ,c,q,r

b− , , , ( f ◦ g))(

g−1(

g(a)m

); p)

2

≤ m+1

2

⎡⎣ f (g(a))−m2 f

(g(a)m2

)mg(b)−g(a)

(gmw′, ,c,q,r

b−, , , 1)(

g−1(

g(a)m

); p

)

+(

f (g(b))+mf

(g(a)m2

))(gmw′, ,c,q,r

b−, , , 1)(

g−1(

g(a)m

); p

)],w′ =

w(mg(b)−g(a))

.

Proof. By definition of m-convex function f , we have

2 f

(g(a)+mg(b)

2

)≤ f (tg(a)+m(1− t)g(b))+mf

(tg(b)+ (1− t)

g(a)m

). (7.49)

Further, from (7.49), one can obtain the following integral inequality:

2 f

(g(a)+mg(b)

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)dt (7.50)

≤∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f (tg(a)+m(1− t)g(b))dt

+m∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(tg(b)+ (1− t)

g(a)m

)dt.

Setting g(x) = tg(a)+ m(1− t)g(b) and g(y) = tg(b)+ (1− t) g(a)m in (7.50), we get the

following inequality:

2 f

(g(a)+mg(b)

2

)(g

,r,k,c ,, ,w′,a+1

)(g−1(mg(b)); p) (7.51)

≤(

g ,r,k,c ,, ,w′,a+( f ◦ g)

)(g−1(mg(b)); p)+m+1

(gmw′, ,c,q,r

b−, , , ( f ◦ g))(

g−1(

g(a)m

); p

).


Also by using the m-convexity of f , we obtain

f (tg(a)+m(1− t)g(b))+mf

(tg(b)+ (1− t)

g(a)m

)(7.52)

≤ m

(f (g(b))+mf

(g(a)m2

))+(

f (g(a))−m2 f

(g(a)m2

))t.

This leads to the following integral inequality:∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f (tg(a)+m(1− t)g(b))dt (7.53)

+m∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(tg(b)+ (1− t)

g(a)m

)dt

≤ m

(f (g(b))+mf

(g(a)m2

))∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)dt

+(

f (g(a))−m2 f

(g(a)m2

))∫ 1

0tE ,c,q,r

, ,, (wt ; p)dt.

Again by setting g(x) = tg(a)+m(1− t)g(b), g(y) = tg(b)+(1− t) g(a)m in (7.53) and after

calculation, we get(g

,r,k,c ,, ,w′,a+( f ◦ g)

)(g−1(mg(b)); p)+m+1

(gmw′, ,c,q,r

b−, , , ( f ◦ g))(

g−1(

g(a)m

); p

)

(7.54)

≤ m+1

⎛⎝ f (g(a))−m2 f

(g(a)m2

)mg(b)−g(a)

(gmw′, ,c,q,r

b−, , , 1)(

g−1(

g(a)m

); p

)

+(

f (g(b))+mf

(g(a)m2

))(gmw′, ,c,q,r

b−, , , 1)(

g−1(

g(a)m

); p

)).

By combining (7.51) and (7.54), we get the desired result. �

Remark 7.9

(i) Setting m = 1 in Theorem 7.14, we obtain [132, Theorem 3.1].

(ii) Setting g = I in Theorem 7.14, we obtain [87, Theorem 3.1].

The following theorem gives the Fejer-Hadamard inequality for m-convex functions.

Theorem 7.15 Let f ,g,h : [a,b]→R, 0 < a < b, Range(g), be such that f is positive andf ∈ L1[a,b], g differentiable and strictly increasing and h integrable and nonnegative. If fis m-convex, m∈ (0,1], g(a) < mg(b) and g(a)+mg(b)−mg(x)= g(x), then the followinginequalities for fractional operator (2.24) hold:

2 f

(g(a)+mg(b)

2

)(gw′, ,c,q,r

mb−, , ,h ◦ g)(

g−1(

g(a)m

); p

)

≤ (1+m)(

gw′, ,c,q,rb−, , , ( f ◦ g)(h ◦ g)

)(g−1

(g(a)m

); p

)


≤f (g(a))−m2 f

(g(a)m2

)(g(b)− g(a)

m )

(gw′, ,c,q,r

b−, , , h ◦ g)(

g−1(

g(a)m

); p

)

+m

(f (g(b))+mf

(g(a)m2

))(gw′, ,c,q,r

b−, , , h ◦ g)(

g−1(

g(a)m

); p

),w′ =

w

(g(b)− g(a)m )

.

Proof. Multiplying both sides of (7.49) by 2t−1h(tg(b)+ (1− t) g(a)

m

)E ,c,q,r , ,, (wt ; p)

and integrating on [0,1], we get

2 f

(g(a)+mg(b)

2

)∫ 1

0t−1h

(tg(b)+ (1− t)

g(a)m

)E ,c,q,r , ,, (wt ; p)dt (7.55)

≤∫ 1

0t−1h

(tg(b)+ (1− t)

g(a)m

)E ,c,q,r , ,, (wt ; p) f (tg(a)+m(1− t)g(b))dt

+m∫ 1

0t−1h

(tg(b)+ (1− t)

g(a)m

)E ,c,q,r , ,, (wt ; p) f

(tg(b)+ (1− t)

g(a)m

)dt.

Setting g(x)= tg(b)+(1−t) g(a)m and also using g(a)+mg(b)−mg(x)= g(x) the following

inequality is obtained:

2 f

(g(a)+mg(b)

2

)(gw′, ,c,q,r

b−, , , h ◦ g)(

g−1(

g(a)m

); p

)(7.56)

≤ (1+m)(

gw′, ,c,q,rb−, , , ( f ◦ g)(h ◦ g)

)(g−1

(g(a)m

); p

).

Multiplying both sides of inequality (7.52) with t−1h(tg(b)+ (1− t) g(a)

m

)E ,c,q,r , ,, (wt ; p)

and integrating on [0,1], then setting g(x) = tg(b) + (1− t) g(a)m and also using g(a) +

mg(b)−mg(x) = g(x) we have

(1+m)(

gw′, ,c,q,rb−, , , ( f ◦ g)(h ◦ g)

)(g−1

(g(a)m

); p

)(7.57)

≤( f (g(a))−m2 f

(g(a)m2

))

g((b)− g(a)m )

(gw′, ,c,q,r

b−, , , h ◦ g)(

g−1(

g(a)m

); p

)

+m

(f (g(b))+mf

(g(a)m2

))(gw′, ,c,q,r

b−, , , h ◦ g)(

g−1(

g(a)m

); p

).

By combining (7.56) and (7.57), we get the desired result. �

Remark 7.10

(i) In the Theorem 7.15, if we put m = 1, then we get [132, Theorem 3.2].

(ii) In the Theorem 7.15, if we put g = I and p = 0, then we get [3, Theorem ].

(iii) In the Theorem 7.15, if we put g = I, then we get [57, Theorem ].


7.4 Hadamard and Fejer-Hadamard Inequalitiesfor (h−m)-convex Functions

In this section the results proved in previous sections are generalized via (h−m)-convexfunctions.

Theorem 7.16 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [0,) → R be an integrable and (h−m)-convex function with m ∈ (0,1]. Then thefollowing inequality for extended generalized fractional integral holds

f

(bm+a

2

)Hw′

a+, (mb; p) (7.58)

≤ h

(12

)[m+1(w′m , ,c,q,r

b−, , , f )( a

m; p)

+(w′, ,c,q,ra+, , , f )(mb; p)

]

≤ h

(12

)(mb−a)

{[m2 f

( am2

)+mf (b)

](w′, ,c,q,r

0+, , , h)(1; p)

+[mf (b)+ f (a)](w′ , ,c,q,r1−, , , h)(0; p)

}where w′ = w

(bm−a) .

Proof. Since f is (h−m)-convex function, we have

f

(xm+ y

2

)≤ h

(12

)(mf (x)+ f (y)). (7.59)

Setting x = (1− t) am + tb and y = m(1− t)b+ ta, t ∈ [0,1], then integrating over [0,1] after

multiply with t−1E ,c,q,r , , (wt ; p) we get

f

(bm+a

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt

≤ h

(12

)[∫ 1

0t−1E ,c,q,r

, , (wt ; p)mf((1− t)

am

+ tb)

dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (m(1− t)b+ ta)dt

].

By substituting s = (1− t) am + tb and z = m(1− t)b+ ta we have

f

(bm+a

2

)Hw′

a+, (mb; p) (7.60)

≤ h

(12

)[m+1(w′, ,c,q,r

b−, , , f )( a

m; p)

+(w′, ,c,q,ra+, , , f )(mb; p)

].

7.4 INEQUALITIES FOR (h−m)-CONVEX FUNCTIONS 101

This completes the proof of first inequality in (7.58). For the second inequality(h−m)-convexity of f also gives

mf((1− t)

am

+ tb)

+ f (m(1− t)b+ ta)

≤ m2h(1− t) f( a

m2

)+mh(t) f (b)+mh(1− t) f (b)+h(t) f (a).

Multiplying both sides of above inequality with h(

12

)t−1E ,c,q,r

, , (wt ; p) and integratingover [0,1], then by change of variables we have

h

(12

)[m+1(w′m , ,c,q,r

b− , , , f )( a

m; p)

+(w′, ,c,q,ra+, , , f )(mb; p)

].

≤ h

(12

)(mb−a)

{[m2 f

( am2

)+mf (b)

]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h(1− t)dt

+[mf (b)+ f (a)]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h(t)dt

}.

This gives us the second inequality in (7.58). �

Several known results are special cases of the above generalized fractional Hadamardinequality shown in the following remark.

Remark 7.11


(ii) If h(t) = t, p = 0 and m = 1, then we get [51, Theorem 2.1].

(iii) If h(t) = t, p = 0, then we get [50, Theorem 3].

(iv) If h(t) = t, p = 0 and w = 0, then we get [65, Theorem 2.1].

(v) If h(t) = t, p = 0, m = 1 and w = 0, then we get [140, Theorem 2].

(vi) If h(t) = t, p = 0, m = 1, = 1 and w = 0, then we get the Hadamard inequality.

Another generalized Hadamard inequality containing Mittag-Leffler function is given inthe following theorem.

Theorem 7.17 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [0,) → R be an integrable and (h−m)-convex function with m ∈ (0,1]. Then thefollowing inequality for extended generalized fractional integral holds

f

(a+bm

2

)Hw′2

( a+bm2 )+

,(mb; p) (7.61)

≤ h

(12

)[(w′2 , ,c,q,r

( a+bm2 )+

, , ,f )(mb; p)+m+1(w′(2m) , ,c,q,r

( a+bm2m )−, , ,

f )( a

m; p)]

≤ h

(12

)(mb−a)

2

{(m2 f

( am2

)+mf (b)

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)h(

2− t2

)dt

+[mf (b)+ f (a)]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h( t

2

)dt

}.


Proof. Putting x = t2b+ (2−t)

2am and y = t

2a+m (2−t)2 b in (7.59), we get

f

(a+bm

2

)≤ h

(12

)[mf

(t2b+

(2− t)2

am

)+ f

(t2a+m

(2− t)2

b

)]. (7.62)

Multiplying (7.62) by t−1E ,c,q,r , , (wt ; p) on both sides, then integrating over [0,1], we

have

f

(a+bm

2

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt

≤ h

(12

)[∫ 1

0t−1E ,c,q,r

, , (wt ; p)mf

(t2a+m

(2− t)2

b

)dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(t2b+

(2− t)2

am

)dt

].

By substituting x = t2b+ (2−t)

2am and y = t

2a+m (2−t)2 b, one gets

f

(a+bm

2

)Hw′2

( a+bm2 )+

,(mb; p) (7.63)

≤ h

(12

)[(w′2 , ,c,q,r

( a+bm2 )+

, , ,f )(mb; p)++m+1(w′(2m) , ,c,q,r

( a+bm2m )−, , ,

f )( a

m; p)]

.

Again by using (h−m)−convexity of f , we have

f

(t2a+m

(2− t)2

b

)+mf

(t2b+

(2− t)2

am

)(7.64)

≤ h( t

2

)f (a)+mh

(2− t

2

)f (b)+mh

( t2

)f (b)+m2h

(2− t

2

)f( a

m2

)

= m[mf

( am2

)+ f (b)

]h

(2− t

2

)+[mf (b)+ f (a)]h

( t2

).

Multiplying (7.64) by h(1

2

)t−1E ,c,q,r

, , (wt ; p) on both sides, then integrating over [0,1],we have

h

(12

)[∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(t2a+m

(2− t)2

b

)dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p)mf

(t2b+

(2− t)2

am

)dt

]

≤ h

(12

){m[mf

( am2

)+ f (b)

]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h(

2− t2

)dt

+[mf (b)+ f (a)]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h( t

2

)dt

}.

By using change of variables we conclude

h

(12

)[(w′2 , ,c,q,r

( a+bm2 )+

, , ,f )(mb; p)++m+1(w′(2m) , ,c,q,r

( a+bm2m )−, , ,

f )( a

m; p)]

7.4 INEQUALITIES FOR (h−m)-CONVEX FUNCTIONS 103

≤ h

(12

)(bm−a)

2

{m[mf

( am2

)+ f (b)

]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h(

2− t2

)dt

+[mf (b)+ f (a)]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h( t

2

)dt

}.

From above inequality and (7.63), we get the required inequality (7.61). �

Remark 7.12

(i) If we put p = 0 in (7.61), then [131, Theorem 2.2] is obtained.

(ii) If h(t) = t, p = 0 and m = 1, then we get [131, Corollary 2.3].

(iii) If h(t) = t, then we get Theorem 7.12.

(iv) If h(t) = t, p = 0 and w = 0, then we get Theorem 7.2.

(v) If m = 1, then we get the inequality for h-convex function.

Theorem 7.18 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R be a (h−m)-convex function with 0 ≤ a < b and f ∈ L1[a,b]. Also, letg : [a,b]→ R be a function which is nonnegative, integrable and symmetric about a+mb

2 . Iff (mb+a−mx) = f (x), then the following inequalities for extended generalized fractionalintegrals hold

f

(bm+a

2

)(w′ , ,c,q,r

b− , , , g)( a

m; p)≤ h

(12

)(m+1)(w′, ,c,q,r

b−, , , f g)( a

m; p)

(7.65)

≤ h

(12

){[m2 f

( am2

)+mf (b)

](w′ , ,c,q,r

0+, , , h)(1; p)

+[mf (b)+ f (a)](w′ , ,c,q,r1− , , , h)(0; p)

},

where w′ = w(bm−a) .

Proof. Since f is (h−m)-convex function, for t ∈ [a,b] we have

f

(bm+a

2

)≤ h

(12

)[mf

((1− t)

am

+ tb)

+ f (m(1− t)b+ ta)].

Multiplying both sides of above inequality with t−1E ,c,q,r , , (wt ; p)g

(tb+(1− t) a

m

)and

integrating over [0,1], we have

f

(bm+a

2

)∫ 1

0t−1E ,c,,r

, ,(wt ; p)g(tb+(1− t)

am

)dt

≤ h

(12

)[∫ 1

0t−1E ,c,q,r

, , (wt ; p)g(tb+(1− t)

am

)mf

((1− t)

am

+ tb)

dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p)g(tb+(1− t)

am

)f (m(1− t)b+ ta)dt

].


If we set x = (1−t) am +tb and use the given condition f (mb+a−mx)= f (x), we conclude

f

(bm+a

2

)(w′, ,c,q,r

b−, , , g)( a

m; p)≤ h

(12

)(m+1)(w′, ,c,q,r

b−, , , f g)( a

m; p)

. (7.66)

This completes the proof of first inequality in (7.65). For the second inequality using(h−m)-convexity of f we have

mf((1− t)

am

+ tb)

+ f (m(1− t)b+ ta)

≤ m2h(1− t) f( a

m2

)+mh(t) f (b)+mh(1− t) f (b)+h(t) f (a).

Multiplying both sides with h(

12

)t−1E ,c,q,r

, , (wt ; p)g(tb+(1− t) a

m

)and integrating over

[0,1], we have

h

(12

)(m+1)(w′, ,c,q,r

b−, , , f g)( a

m; p)

≤h

(12

){[m2 f

( am2

)+mf (b)

]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h(1− t)g(tb+(1− t)

am

)dt

+[mf (b)+ f (a)]∫ 1

0t−1E ,c,q,r

, , (wt ; p)h(t)g(tb+(1− t)

am

)dt

}.

By change of variables the second inequality in (7.65) can be obtained. �

Remark 7.13

(i) If we put p = 0 in (7.65), then [131, Theorem 2.5] is obtained.

(ii) If we put h(t) = t, m = 1 and p = 0 in (7.65), then [51, Theorem 2.2] is obtained.

7.5 Hadamard and Fejer-Hadamard inequalitiesfor Harmonically Convex Functions

In the following we give the Hadamard inequality for harmonically convex functions.

Theorem 7.19 [76] Let f : I ⊂R\{0}→R be harmonically convex function and a,b∈ Iwith a < b. If f ∈ L[a,b] then the following inequality hold:

f

(2aba+b

)≤ ab

b−a

∫ b

a

f (x)x2 dx ≤ f (a)+ f (b)

2. (7.67)

A Fejer-Hadamard inequality for harmonically convex functions is stated as follows.

7.5 INEQUALITIES FOR HARMONICALLY CONVEX FUNCTIONS 105

Theorem 7.20 [33] Let f : I ⊂ R \ {0} → R be a harmonically convex function anda,b∈ I with a < b. If f ∈ L[a,b] and Let u : [a,b]⊂R\{0}→R is non negative integrableand harmonically symmetric with respect to 2ab

a+b then the following inequality for fractionalintegral operator hold:

f

(2aba+b

)∫ b

a

u(x)x2 dx ≤

∫ b

a

f (x)u(x)x2 dx ≤ f (a)+ f (b)

2

∫ b

a

u(x)x2 . (7.68)

A version of the Fejer-Hadamard inequality for harmonically convex functions via Riemann-Liouville fractional integrals is stated as follows.

Theorem 7.21 [79] Let f : I ⊂ (0,) → R be a function such that a,b ∈ I witha < b. If f ∈ L[a,b] and f is harmonically convex function then the following inequal-ity for fractional integral operator hold:

f

(2aba+b

)≤ ( +1)

2

(ab

b−a

) [J1

b+( f ◦ h)(

1a)+ J1

a−( f ◦ h)(

1b)]≤ f (a)+ f (b)

2(7.69)

where h(x) = 1x and x ∈ [a,b].

Another version of the Fejer-Hadamard inequality for harmonically convex functions viaRiemann-Liouville fractional integrals is stated as follows.

Theorem 7.22 [95] Let f : I ⊂ (0,) → R be a harmonically convex function on [a,b]for a,b ∈ I with a < b. If f ∈ L[a,b] then the following inequalities for fractional integralhold

f

(2aba+b

)≤ ( +1)

21−

(ab

b−a

) (Ja+b

2ab −f ◦ g

(1b

)+ Ja+b

2ab+ f ◦ g

(1a

))

≤ f (a)+ f (b)2

where g(t) = 1t for t ∈ [ 1

b , 1a ].

The following Fejer-Hadamard inequality for harmonically convex functions via extendedMittag-Leffler functions is the generalized version.

Theorem 7.23 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] ⊂ (0,) → R be such that f ∈ L1[a,b] with a < b. If f is a harmonically con-vex function on [a,b], then the following inequalities for extended generalized fractionalintegral operators hold

f

(2aba+b

)Hw′

1a−,

(1b; p

)

≤ 12

((w′, ,c,q,r

1a−, , ,

f ◦ g

)(1b; p

)+(w′, ,c,q,r

1b

+, , ,

f ◦ g

)(1a; p

))

≤ f (a)+ f (b)2

Hw′1b

+,

(1a; p

),

where w′ = w( abb−a ) and g(t) = 1

t for t ∈ [ 1b , 1

a ].


Proof. Since f is harmonically convex function, for all x,y ∈ [a,b], f(

2xyx+y

)≤ f (x)+ f (y)

2 .

For x = abtb+(1−t)a and y = ab

ta+(1−t)b we have

2 f

(2aba+b

)≤ f

(ab

tb+(1− t)a

)+ f

(ab

ta+(1− t)b

). (7.70)

Multiplying (7.70) by t−1E ,c,q,r , , (wt ; p) on both sides, and integrating over [0,1] we have

2 f

(2aba+b

)∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt ≤∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(ab

tb+(1− t)a

)dt

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(ab

ta+(1− t)b

)dt.

If we put in above x = tb+(1−t)aab and y = ta+(1−t)b

ab , then we have the following inequality

2 f

(2aba+b

)∫ 1a

1b

(ab

b−a

)−1(x− 1

b

)−1

(7.71)

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)(ab

b−a

)dx

≤∫ 1

a

1b

(ab

b−a

)−1(x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)

f

(1x

)(ab

b−a

)dx+

∫ 1a

1b

(ab

b−a

)−1(1a− y

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(1a− y

); p

)f

(1y

)(ab

b−a

)dy.

After simplification we get

2 f

(2aba+b

)∫ 1a

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w′(

x− 1b

); p

)dx

≤∫ 1

a

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w′(

x− 1b

); p

)f

(1x

)dx

+∫ 1

a

1b

(1a− y

)−1

E ,c,q,r , ,

(w′(

1a− y

); p

)f

(1y

)dy.

By using Definition 2.2 we get

2 f

(2aba+b

)(w′, ,c,q,r

1a−, , ,

1

)(1b; p

)≤(w′, ,c,q,r

1a−, , ,

f ◦ g

)(1b; p

)

+(w′, ,c,q,r

1b

+, , ,

f ◦ g

)(1a; p

). (7.72)


Again by using that harmonically convexity of f for t ∈ [0,1] one can have

f

(ab

tb+(1− t)a

)+ f

(ab

ta+(1− t)b

)≤ f (a)+ f (b). (7.73)

Multiplying (7.73) by t−1E ,c,q,r , , (wt ; p) on both sides, and integrating over [0,1] we have

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(ab

tb+(1− t)a

)dt (7.74)

+∫ 1

0t−1E ,c,q,r

, , (wt ; p) f

(ab

ta+(1− t)b

)dt

≤ ( f (a)+ f (b))∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt.

By putting in above x = tb+(1−t)aab and y = ta+(1−t)b

ab then after simplifications, we have

(w′, ,c,q,r

1a−, , ,

f ◦ g

)(1b; p

)+(w′, ,c,q,r

1b

+, , ,

f ◦ g

)(1a; p

)

≤ ( f (a)+ f (b))(w′, ,c,q,r

1b

+, , ,

1

)(1a; p

). (7.75)

Inequalities (7.72) and (7.75) provide the required inequality. �

Remark 7.14 In Theorem 7.23,

(i) if we put p = 0, then we get [2, Theorem 3.1],

(ii) if we put w = p = 0, then we get [78, Theorem 4].

Another version is given in the next result.

Theorem 7.24 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b]⊂ (0,)→ R be a harmonically convex function such that f ∈ L1[a,b] with a < b.If f is a harmonically symmetric about a+b

2ab , then the following inequalities for extendedgeneralized fractional integral operators hold

f

(2aba+b

)Hw′

a+b2ab −,

(1b; p

)

≤ 12

((w′, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f ◦ g

)(1b; p

))

≤ f (a)+ f (b)2

Hw′a+b2ab

+,

(1a; p

),

where w′ = w( abb−a ) and g(t) = 1

t , t ∈ [ 1b , 1

a ].


Proof. Multiplying (7.70) by 2t−1E ,c,q,r , , (wt ; p) on both sides, and integrating over

[0, 12 ] we have

2 f

(2aba+b

)∫ 12

0t−1E ,c,q,r

, , (wt ; p)dt ≤∫ 1

2

0t−1E ,c,q,r

, , (wt ; p) f

(ab

ta+(1− t)b

)dt

(7.76)

+∫ 1

2

0t−1E ,c,q,r

, , (wt ; p) f

(ab

tb+(1− t)a

)dt.

Putting in above x = tb+(1−t)aab that is ab

ta+(1−t)b = 11a + 1

b−xthen we have

2 f

(2aba+b

)∫ a+b2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)dx

≤∫ a+b

2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)

f

(1

1a + 1

b − x

)dx+

∫ a+b2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

))((

x− 1b

); p

)f

(1x

)dx.

Since f is harmonically symmetric about a+b2ab , we replace 1

a + 1b − x by x in first term on

the right hand side of the above inequality and after simplification we have

2 f

(2aba+b

)∫ a+b2ab

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w′(

x− 1b

); p

)dx

≤∫ 1

a

a+b2ab

(1a− x

)−1

E ,c,q,r , ,

(w′(

1a− x

); p

)f

(1x

)dx

+∫ a+b

2ab

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w′(

x− 1b

); p

)f

(1x

)dx.


2 f

(2aba+b

)(w′, ,c,q,r

a+b2ab −, , ,

1

)(1b; p

)(7.77)

≤(w′, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f ◦ g

)(1b; p

).

Now multiplying (7.73) by t−1E ,c,q,r , , (wt ; p) on both sides, and integrating over [0, 1

2 ]we have ∫ 1

2

0t−1E ,c,q,r

, , (wt ; p) f

(ab

tb+(1− t)a

)dt

+∫ 1

2

0t−1E ,c,q,r

, , (wt ; p) f

(ab

ta+(1− t)b

)dt (7.78)


≤ ( f (a)+ f (b))∫ 1

2

0t−1E ,c,q,r

, , (wt ; p)dt.


ta+(1−t)b = 11a + 1

b−xthen we have

∫ a+b2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)f

(1x

)dx

+∫ a+b

2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)

f

(1

1a + 1

b − x

)dx ≤ ( f (a)+ f (b))

∫ a+b2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)dx.

Since f is harmonically symmetric about a+b2ab , using this fact by replacing 1

a + 1b −x with x

in first term of the left hand side of above inequality and after simple calculation, we have(w′, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f ◦ g

)(1b; p

)(7.79)

≤ ( f (a)+ f (b))(

Hw′a+b2ab

+,

)(1a; p

).



(i) if p = 0, then we get [2, Theorem 3.3],

(ii) if w = p = 0, then we get Theorem 7.22.

The following lemma is needed to prove the next result.

Lemma 7.1 Let f : [a,b]⊂R\{0}→R be integrable and harmonically symmetric abouta+b2ab . Then we have the following equality for extended generalized fractional integraloperators holds(

w, ,c,q,ra+b2ab

+, , ,

f ◦ g

)(1a; p

)=(w, ,c,q,r

a+b2ab

−, , ,

f ◦ g

)(1b; p

)(7.80)

=12

((w, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)+(w, ,c,q,r

a+b2ab

−, , ,

f ◦ g

)(1b; p

))

where g(t) = 1t for t ∈ [ 1

b , 1a ].


Proof. Since f is harmonically symmetric about a+b2ab , we have f ( 1

x ) = f

(1

1a + 1

b−x

). By

the definition of extended generalized fractional integral operator,(w, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)=∫ 1

a

a+b2ab

(1a− t

)−1

E ,c,q,r , ,

(w

(1a− t

); p

)f

(1t

)dt,

(7.81)replace t by 1

a + 1b − x in above we have(

w, ,c,q,ra+b2ab

+, , ,

f ◦ g

)(1a; p

)

=∫ a+b

2ab

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w

(x− 1

b

); p

)f

(1

1a + 1

b − x

)dx

=∫ a+b

2ab

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w

(x− 1

b

); p

)f

(1x

)dx.

By using Definition 2.2 we get(w, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)=(w, ,c,q,r

a+b2ab

−, , ,

f ◦ g

)(1b; p

). (7.82)

By adding

(w, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)( 1

a ; p) to both sides of above we have

2

(w, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)=(w, ,c,q,r

a+b2ab

+, , ,

f ◦ g

)(1a; p

)(7.83)

+(w, ,c,q,r

a+b2ab

−, , ,

f ◦ g

)(1b; p

)

which is required. �

The next result provides the generalized version of Fejer-Hadamard inequality for har-monically convex functions.

Theorem 7.25 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] ⊂ (0,) → R be a harmonically convex function with a < b. Let f ∈ L1[a,b] andalso let g : [a,b]→ R be a nonnegative, integrable and harmonically symmetric about a+b

2ab .Then the following inequalities for extended generalized fractional integral operators hold

f

(2aba+b

)[(w′, ,c,q,r

a+b2ab

+, , ,

g ◦ h

)(1a; p

)+(w′ , ,c,q,r

a+b2ab −, , ,

g ◦ h

)(1b; p

)](7.84)

≤(w′ , ,c,q,r

a+b2ab

+, , ,

f g ◦ h

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f g ◦ h

)(1b; p

)

≤ f (a)+ f (b)2

[(w′, ,c,q,r

a+b2ab

+, , ,

g ◦ h

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

g ◦ h

)(1b; p

)],

where w′ = w( abb−a) and h(t) = 1

t for t ∈ [ 1b , 1

a ].


Proof. Multiplying (7.70) by t−1E ,c,q,r , , (wt ; p)g

(ab

tb+(1−t)a

)on both sides, and inte-

grating over [0, 12 ] we have

2 f

(2aba+b

)∫ 12

0t−1E ,c,q,r

, , (wt ; p)g(

abtb+(1− t)a

)dt (7.85)

≤∫ 1

2

0t−1E ,c,q,r

, , (wt ; p) f

(ab

ta+(1− t)b

)g

(ab

tb+(1− t)a

)dt

+∫ 1

2

0t−1E ,c,q,r

, , (wt ; p) f

(ab

tb+(1− t)a

)g

(ab

tb+(1− t)a

)dt.


ta+(1−t)b = 11a + 1

b−xthen we have

2 f

(2aba+b

)∫ a+b2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

) (x− 1

b

); p

)(7.86)

g

(1x

)dx ≤

∫ a+b2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

)(x− 1

b

); p

)

f

(1

1a + 1

b − x

)g

(1x

)dx+

∫ a+b2ab

1b

(ab

b−a

) (x− 1

b

)−1

E ,c,q,r , ,

(w

(ab

b−a

))((

x− 1b

); p

)f

(1x

)g

(1x

)dx.

Since f is harmonically symmetric about a+b2ab , after simplification (7.86) becomes

2 f

(2aba+b

)∫ a+b2ab

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w′(

x− 1b

); p

)g

(1x

)dx (7.87)

≤∫ 1

a

a+b2ab

(1a− x

)−1

E ,c,q,r , ,

(w′(

1a− x

); p

)f

(1x

)g

(1x

)dx

+∫ a+b

2ab

1b

(x− 1

b

)−1

E ,c,q,r , ,

(w′(

x− 1b

); p

)f

(1x

)g

(1x

)dx.


2 f

(2aba+b

)(w′, ,c,q,r

a+b2ab −, , ,

g ◦ h

)(1b; p

)

≤(w′, ,c,q,r

a+b2ab

+, , ,

f g ◦ h

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f g ◦ h

)(1b; p

).

Using Lemma 7.1 in above inequality, we have

f

(2aba+b

)[(w′, ,c,q,r

a+b2ab −, , ,

g ◦ h

)(1b; p

)+(w′, ,c,q,r

a+b2ab

+, , ,

g ◦ h

)(1a; p

)](7.88)

≤(w′, ,c,q,r

a+b2ab

+, , ,

f g ◦ h

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f g ◦ h

)(1b; p

).


Now multiplying (7.73) by t−1E ,c,q,r , , (wt ; p)g

(ab

tb+(1−t)a

)on both sides, and integrating

over [0, 12 ] we have

∫ 12

0t−1E ,c,q,r

, , (wt ; p) f

(ab

tb+(1− t)a

)g

(ab

tb+(1− t)a

)dt (7.89)

+∫ 1

2

0t−1E ,c,q,r

, , (wt ; p) f

(ab

ta+(1− t)b

)g

(ab

tb+(1− t)a

)dt

≤ ( f (a)+ f (b))∫ 1

2

0t−1E ,c,q,r

, , (wt ; p)g(

abtb+(1− t)a

)dt.

By putting x = tb+(1−t)aab and using harmonically symmetry of f with respect to a+b

2ab inabove then after simplification we have(

w′ , ,c,q,ra+b2ab

+, , ,

f g ◦ h

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f g ◦ h

)(1b; p

)(7.90)

≤ ( f (a)+ f (b))(w′, ,c,q,r

a+b2ab

+, , ,

g ◦ h

)(1a; p

).

Using Lemma 7.1 in (7.90) we have(w′, ,c,q,r

a+b2ab

+, , ,

f g ◦ h

)(1a; p

)+(w′, ,c,q,r

a+b2ab −, , ,

f g ◦ h

)(1b; p

)(7.91)

≤ ( f (a)+ f (b))2

(w′, ,c,q,r

a+b2ab −, , ,

g ◦ h

)(1b; p

)+(w′, ,c,q,r

a+b2ab

+, , ,

g ◦ h

)(1a; p

).



(i) if p = 0, then we get [2, Theorem 3.6].

(ii) if w = p = 0 and g(x) = 1, then we get Theorem 7.22.

Corollary 7.1 In Theorem 7.25, if we put w = p = 0, then we get the following inequali-ties via Riemann-Liouville fractional integral operator

f

(2aba+b

)[(Ja+b

2ab+g ◦ h

)(1a

)+(

Ja+b2ab −

g ◦ h

)(1b

)]

≤(

Ja+b2ab

+ f g ◦ h

)(1a

)+(

Ja+b2ab −

f g ◦ h

)(1b

)

≤ f (a)+ f (b)2

[(Ja+b

2abb+g ◦ h

)(1a

)+(

Ja+b2ab −

g ◦ h

)(1b

)].

In the next results generalized fractional integral operators by a monotone increasing func-tion are utilized.


Theorem 7.26 Let f ,g : [a,b]→R, 0 < a < b, Range(g)⊂ [a,b] be such that f is positiveand f ∈ L1[a,b] and g differentiable and strictly increasing. If f is a harmonically convexfunction on [a,b], then for fractional integral operators (2.23) and (2.24) we have:

f

(2g(a)g(b)g(a)+g(b)

)(gw′. ,c,q,r(

g−1( 1g(a) )

)−, , ,

1

)(g−1

(1

g(b)

); p

)(7.92)

≤ 12

((gw′. ,c,q,r(

g−1( 1g(a) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

)

+

(gw′ . ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

))

≤ f (g(a))+ f (g(b))2

(gw′ . ,c,q,r(

g−1( 1g(b) )

)+, , ,

1

)(g−1

(1

g(a)

); p

),

where (t) = 1g(t) for all t ∈ [ 1

b , 1a ] and w′ = w

(g(a)(b)

g(b)−g(a)

).

Proof. Since f is harmonically convex on [a,b], for x,y ∈ [a,b], the following inequalityholds:

f

(2g(x)g(y)g(x)+g(y)

)≤ f (g(x))+ f (g(y))

2. (7.93)

By taking g(x) = g(a)g(b)tg(b)+(1−t)g(a) and g(y) = g(a)g(b)

tg(a)+(1−t)g(b) in (7.93), we have

2 f

(2g(a)g(b)g(a)+g(b)

)≤ f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)+ f

(g(a)g(b)

tg(a)+ (1− t)g(b)

). (7.94)

Multiplying (7.94) by t−1E ,c,q,r , ,, (w(t); p), and integrating over [0,1] we get

2 f

(2g(a)g(b)g(a)+g(b)

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)dt (7.95)

≤∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt

+∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)dt.

By setting g(x) = tg(b)+(1−t)g(a)g(a)g(b) and g(y) = tg(a)+(1−t)g(b)

g(a)g(b) in (7.95) and using (2.23),(2.24), the first inequality of (7.92) can be obtained. On the other hand from harmoni-cally convexity of f we have

f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)+ f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)≤ f (g(a))+ f (g(b)). (7.96)

Multiplying (7.96) by t−1E ,c,q,r , ,, (w(t); p) and then integrating over [0,1] we get∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt (7.97)


+∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)dt

≤ ( f (g(a))+ f (g(b)))∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)dt.

By setting g(x) = tg(b)+(1−t)g(a)g(a)g(b) and g(y) = tg(a)+(1−t)g(b)

g(a)g(b) in (7.97), and using (2.23),(2.24), the second inequality of (7.92) can be obtained. �

Remark 7.17

(i) By setting p = 0 and g = I, [2, Theorem 3.1] is obtained.

(ii) By setting g = I, [64, Theorem 2.1] is obtained.

(iii) By setting w = p = 0, g = I, [78, Theorem 4] is obtained.

Corollary 7.2 If we take (x) = x in Theorem 7.26, then we get the following inequali-ties.

f

(2

a+b

)(gw′. ,c,q,r

(g−1( 1a ))−, , ,

1

)(1b; p

)

≤ 12

((gw′ . ,c,q,r

(g−1( 1a ))−, , ,

f

)(1b; p

)+(

gw′ . ,c,q,r

(g−1( 1b ))+

, , ,f

)(1a; p

))

≤ f ( 1a )+ f ( 1

b )2

(gw′. ,c,q,r

(g−1( 1b ))+

, , ,1

)(1a; p

),

where g is reciprocal function.

The following lemma is useful to give the next result.

Lemma 7.2 Let f ,g : [a,b] → R, 0 < a < b, Range(g) ⊂ [a,b] be the functions suchthat f be positive and f ∈ L1[a,b] and g be a differentiable and strictly increasing. If

f(

1g(x)

)= f

(1

1g(a) + 1

g(b)−g(x)

), then for operators (2.23) and (2.24) we have:

(gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.98)

=

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

)

=12

((gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)

+

(g

,r,k,c

,, ,w,(g−1( 1

g(a) ))− f ◦

)(g−1

(1

g(b)

); p

)),


b , 1a ].


Proof. By operators (2.23) and (2.24), we can write

(gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.99)

=∫ g−1( 1

g(a) )

g−1( 1g(b) )

(1

g(a)−g(t)

)−1

E ,c,q,r , ,,

(w

(1

g(a)−g(t)

); p

)f

(1

g(t)

)d(g(t)).

Replacing g(t) by 1g(a) + 1

g(b) −g(x) in equation (7.99) and then using f(

1g(x)

)= f

(1

1g(a) +

1g(b)−g(x)

), we have

(gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.100)

=

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

).

By adding

(gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p)

on both sides of (7.100), we have

2

(gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.101)

=

(gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)

+

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

)

The equations (7.100) and (7.101) give required result. �

Theorem 7.27 Let f ,g,h : [a,b] → R, 0 < a < b, Range(g),Range(h) ⊂ [a,b] be suchthat f is positive and f ∈ L1[a,b], g differentiable, strictly increasing and h nonnegative

and integrable. If f is harmonically convex and f(

1g(x)

)= f

(1

1g(a) + 1

g(b)−g(x)

), then for

fractional integral operators (2.23) and (2.24) we have:


f

(2g(a)g(b)g(a)+g(b)

)((gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

h ◦)(

g−1(

1g(a)

); p

)(7.102)

+

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

h ◦)(

g−1(

1g(b)

); p

))

≤(

gw. ,c,q,r(g−1( 1

g(b) ))+

, , ,f h ◦

)(g−1

(1

g(a)

); p

)

+

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

f h ◦)(

g−1(

1g(b)

); p

)

≤ f (g(a))+ f (g(b))2

((gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

h ◦)(

g−1(

1g(a)

); p

)

+

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

h ◦)(

g−1(

1g(b)

); p

)),


b , 1a ], f h ◦ = ( f ◦)(h ◦) and w′ = w

(g(a)g(b)

g(b)−g(a)

).

Proof. Multiplying (7.94) by t−1E ,c,q,r , ,, (wt ; p)h

(g(a)g(b)

tg(b)+(1−t)g(a)

), then integrating over

[0,1] we get

2 f

(2g(a)g(b)g(a)+g(b)

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h(

g(a)g(b)tg(b)+ (1− t)g(a)

)dt (7.103)

≤∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt

+∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt.

By setting g(x) = tg(b)+(1−t)g(a)g(a)g(b) in (7.103) and using (2.23), (2.24), and the condition

f(

1g(x)

)= f

(1

1g(a) + 1

g(b)−g(x)

), we have

2 f

(2g(a)g(b)g(a)+g(b)

)(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

h ◦)(

g−1(

1g(b)

); p

)(7.104)

≤(

gw. ,c,q,r(g−1( 1

g(b) ))+

, , ,f h ◦

)(g−1

(1

g(a)

); p

)

+

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

f h ◦)(

g−1(

1g(b)

); p

).

By using Lemma 7.2 in above inequality, we get first inequality in (7.102). For second

inequality of (7.102), multiplying (7.96) by t−1E ,c,q,r , ,, (wt ; p)h

(g(a)g(b)

tg(b)+(1−t)g(a)

), then


integrating over [0,1] we get

∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt (7.105)

+∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt

≤ ( f (g(a))+ f (g(b)))∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h(


)dt.

Setting g(x)= tg(b)+(1−t)g(a)g(a)g(b) in (7.105) and using (2.23), (2.24), and the condition f

(1

g(x)

)=

f

(1

1g(a) +

1g(b)−g(x)

), we have

(g

,r,k,c

,, ,w′,(g−1( 1

g(b) ))+ f h ◦

)(g−1

(1

g(a)

); p

)(7.106)

+

(gw. ,c,q,r(

g−1( 1g(a) )

)−, , ,

f h ◦)(

g−1(

1g(b)

); p

)

≤ ( f (g(a))+ f (g(b)))

(gw. ,c,q,r(

g−1( 1g(b) )

)+, , ,

h ◦)(

g−1(

1g(a)

); p

).

Again using Lemma 7.2 in (7.106), we get the second inequality of (7.102). �

Remark 7.18

(i) By setting p = 0, h(x) = 1 and g = I, [2, Theorem 3.1] is obtained.

(ii) By setting g = I and h(x) = 1, [64, Theorem 2.1] is obtained.

(iii) By setting w = p = 0, h(x) = 1 and g = I, [78, Theorem 4] is obtained.

(iv) By setting w = p = 0, = 1 and g = I, [33, Theorem 8] is obtained.

(v) By setting w = p = 0, = 1, h(x) = 1 and g = I, [95, Theorem 2.4] is obtained.

Theorem 7.28 Let f ,g : [a,b]→R, 0 < a < b, Range(g)⊂ [a,b] be such that f is positiveand f ∈ L1[a,b] and g differentiable and strictly increasing. If f is harmonically convex

on [a,b] and f(

1g(x)

)= f

(1

1g(a) + 1

g(b)−g(x)

), then for operators (2.23) and (2.24) we have:

f

(2g(a)g(b)g(a)+g(b)

)(gw. ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

1

)(g−1

(1

g(b)

); p

)(7.107)

≤ 12

((gw′ , ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)


+

(gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

))

≤ f (g(a))+ f (g(b))2

(gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

1

)(g−1

(1

g(a)

); p

),

where (t) = 1g(t) for t ∈ [ 1

b , 1a ] and w′ = w

(g(a)g(b)

g(b)−g(a)

).

Proof. Multiplying (7.94) by 2t−1E ,c,q,r , ,, (wt ; p) then integrating over [0, 1

2 ] we have

2 f

(2g(a)g(b)g(a)+g(b)

)∫ 12

0t−1E ,c,q,r

, ,, (wt ; p)dt (7.108)

≤∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)dt

+∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt.

Setting g(x) = tg(b)+(1−t)g(a)g(a)g(b) in (7.108) and using f

(1

g(x)

)= f

(1

1g(a) +

1g(b)−g(x)

), (2.23)

and (2.24) the first inequality of (7.107) can be obtained.For second inequality multiplying (7.96) by t−1E ,c,q,r

, ,, (wt ; p) then integrating over[0, 1

2 ], we get

∫ 12

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt (7.109)

+∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)dt

≤ ( f (g(a))+ f (g(b)))∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p)dt.

Setting g(x) = tg(b)+(1−t)g(a)g(a)g(b) in (7.109) and using f

(1

g(x)

)= f

(1

1g(a) +

1g(b)−g(x)

), (2.23)

and (2.24), the second inequality of (7.107) can be obtained. �

Remark 7.19



(iii) By setting w = p = 0 and g = I, [95, Theorem 4] is obtained.

To prove the next result the following lemma is needed:


Lemma 7.3 Let f ,g : [a,b] → R, 0 < a < b, Range(g) ⊂ [a,b] be the functions suchthat f be positive and f ∈ L1[a,b] and g be a differentiable and strictly increasing. If f

is a harmonically convex and f(

1g(x)

)= f

(1

1g(a) +

1g(b)−g(x)

), then for fractional integral

operators (2.23) and (2.24) we have:

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.110)

=

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

)

=12

((gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)

+

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

)), (t) =

1g(t)

,t ∈[1b,1a

].

Proof. By using Definition 7.65, we can write

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.111)

=∫ g−1( 1

g(a) )

g−1( g(a)+g(b)2g(a)g(b) )

(1

g(a)−g(t))−1E ,c,q,r

, ,, (w(1

g(a)−g(t)) ; p) f (

1g(t)

)d(g(t)).

By replacing g(t) by 1g(a) + 1

g(b) −g(x) in equation (7.111) and using the condition f(

1g(x)

)=

f

(1

1g(a) +

1g(b)−g(x)

), we have

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.112)

=

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

).

By adding

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p)

on both sides of (7.112), we

have

2

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)(7.113)


=

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f ◦)(

g−1(

1g(a)

); p

)

+

(gw, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f ◦)(

g−1(

1g(b)

); p

).

The equations (7.112) and (7.113) give the required result. �

Theorem 7.29 Let f ,g,h : [a,b] → R, 0 < a < b, Range(g),Range(h) ⊂ [a,b] be suchthat f is positive and f ∈ L1[a,b], g differentiable, strictly increasing and h nonnegative

and integrable. If f is harmonically convex and f(

1g(x)

)= f

(1

1g(a) + 1

g(b)−g(x)

), then for

fractional integral operators (2.23) and (2.24) we have

f

(2g(a)g(b)g(a)+g(b)

)((gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

h ◦)(

g−1(

1g(a)

); p

)(7.114)

+

(gw′ , ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

h ◦)(

g−1(

1g(b)

); p

))

≤(

gw′, ,c,q,r(g−1( g(a)+g(b)

2g(a)g(b) ))+

, , ,f h ◦

)(g−1

(1

g(a)

); p

)

+

(gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f h ◦)(

g−1(

1g(b)

); p

)

≤ f (g(a))+ f (g(b))2

((gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

h ◦)(

g−1(

1g(a)

); p

)

+

(gw′ , ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

h ◦)(

g−1(

1g(b)

); p

)),

where (t) = 1g(t) for t ∈ [ 1

b , 1a ], f h ◦ = ( f ◦)(h ◦) and w′ = w

(g(a)g(b)

g(b)−g(a)

).

Proof. Multiplying (7.94) by t−1E ,c,q,r , ,, (wt ; p)h

(g(a)g(b)

tg(b)+(1−t)g(a)

)then integrating over

[0, 12 ], we have

2 f

(2g(a)g(b)g(a)+g(b)

)∫ 12

0t−1E ,c,q,r

, ,, (wt ; p)h(


)dt (7.115)

≤∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt

+∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt.


By choosing g(x) = tg(b)+(1−t)g(a)g(a)g(b) and using the condition f

(1

g(x)

)= f

(1

1g(a) + 1

g(b)−g(x)

),

in (7.115) we have

2 f

(2g(a)g(b)g(a)+g(b)

)(gw′ , ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

h ◦)(

g−1(

1g(b)

); p

)(7.116)

≤(

gw′ , ,c,q,r(g−1( g(a)+g(b)

2g(a)g(b) ))+

, , ,f h ◦

)(g−1

(1

g(a)

); p

)

+

(gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f h ◦)(

g−1(

1g(b)

); p

).

Using Lemma 7.3 in above inequality we obtain the first inequality of (7.114).For second inequality of (7.114), multiplying (7.96) by

t−1E ,c,q,r , ,, (wt ; p)h

(g(a)g(b)

tg(b)+(1−t)g(a)

)then integrating over [0, 1

2 ], we have

∫ 12

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(b)+ (1− t)g(a)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt (7.117)

+∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p) f

(g(a)g(b)

tg(a)+ (1− t)g(b)

)h

(g(a)g(b)

tg(b)+ (1− t)g(a)

)dt

≤ ( f (g(a))+ f (g(b)))∫ 1

2

0t−1E ,c,q,r

, ,, (wt ; p)h(


)dt.

Setting g(x) = tg(b)+(1−t)g(a)g(a)g(b) in (7.117) and using (2.23), (2.24) and condition f

(1

g(x)

)=

f

(1

1g(a) +

1g(b)−g(x)

), we have

(gw′ , ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

f h ◦)(

g−1(

1g(a)

); p

)(7.118)

+

(gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)−, , ,

f h ◦)(

g−1(

1g(b)

); p

)

≤ ( f (g(a))+ f (g(b)))

(gw′, ,c,q,r(

g−1( g(a)+g(b)2g(a)g(b) )

)+, , ,

h ◦)(

g−1(

1g(a)

); p

).

Again using Lemma 7.3 in (7.118), the second inequality of (7.114) can be obtained. �

Remark 7.20



(iii) By setting w = p = 0, g = I and = 1, [33, Theorem 8], is obtained.


Corollary 7.3 When we set w = p = 0 and g = I in Theorem 7.29, then we get the fol-lowing inequalities via Riemann-Liouville fractional integrals:

f

(2aba+b

)((Ia+b

2ab+h ◦

)(1a

)+(

Ia+b2ab

−h ◦)(

1b

))

≤(

Ia+b2ab

+ f h ◦)(

1a

)+(

Ia+b2ab

− f h ◦)(

1b

)

≤ f (a)+ f (b)2

((Ia+b

2ab+h ◦

)(1a

)+(

Ia+b2ab

−h ◦)(

1b

)).

7.6 Hadamard and Fejer-Hadamard Inequalitiesfor Harmonically (,h−m)-convex Functions

In this whole section the following notations are frequently used:(F a+

b,

)(w, f )=

(gw, ,c,q,r

a+, , , f)

(b; p),(F b−

a,

)(w, f )=

(gw, ,c,q,r

b−, , , f)

(a; p).

Theorem 7.30 Let f ,g : [a,b]→R, 0 < a < b, Range(g)⊂ [a,b] be such that f is positiveand f ∈ L1[a,b] and g differentiable and strictly increasing. If f is harmonically (,h−m)-convex on [a,b], then for operators (2.23) and (2.24) we have:

f

(2m2g(a)g(b)g(a)+mg(b)

)⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,1) (7.119)

≤ h

(12

)⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′, f ◦)

+m1−2h

(1− 1

2

)(F

g−1(

1mg(b)

)+

g−1(

1g(a)

),

)(w′, f ◦) ≤

(mg(b)−g(a)m2g(a)g(b)

)

{m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h(t)dt

+(

h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h(1− t)dt

},

where w′ = w(

mg(a)(b)mg(b)−g(a)

)and (t) = 1

g(t) for all t ∈ [ 1b , 1

a ].

Proof. Since f is harmonically (,h−m)-convex, for all x,y ∈ [a,b], the followinginequality holds:

f

(2mg(x)g(y)mg(y)+g(x)

)≤ h

(12

)f (g(x))+mh

(1− 1

2

)f (g(y)). (7.120)

7.6 INEQUALITIES FOR HARMONICALLY (,h−m)-CONVEX FUNCTIONS 123

By setting g(x) = mg(a)g(b)

tg(b)+(1−t) g(a)m

and g(y) = mg(a)g(b)tg(a)+m(1−t)g(b) in (7.120), we have

f


)≤ h

(12

)f

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)(7.121)

+mh

(1− 1

2

)f

(mg(a)g(b)

tg(a)+m(1− t)g(b)

).

Multiplying (7.121) by t−1E ,c,q,r , ,, (w(t ); p), and integrating over [0,1] we get

f


)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)dt (7.122)

≤ h

(12

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)dt

+mh

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

tg(a)+m(1− t)g(b)

)dt.

By setting g(x) = tg(b)+(1−t) g(a)m

mg(a)g(b) and g(y) = tg(a)+m(1−t)g(b)mg(a)g(b) in (7.122) and using (2.23) and

(2.24), the first inequality of (7.119) can be obtained. On the other hand from harmonically(,h−m)-convexity of f we have

h

(12

)f

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)+mh

(1− 1

2

)f

(mg(a)g(b)

tg(a)+m(1− t)g(b)

)(7.123)

≤ h

(12

)(h(1− t) f (g(b))+mh(t) f

(g(a)m2

))

+mh

(1− 1

2

)(h(1− t) f (g(a))+mh(t) f (g(b)))

= m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)h(t)

+(

h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)h(1− t).

Multiplying (7.123) by t−1E ,c,q,r , ,, (w(t ); p) and then integrating over [0,1] we get

h

(12

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)dt (7.124)

+mh

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

tg(a)+m(1− t)g(b)

)dt

≤ m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h(t)dt

+(

h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h(t)h(1− t)dt.



mg(a)g(b) and g(y) = tg(a)+m(1−t)g(b)mg(a)g(b) in (7.124) and using (2.23) and

(2.24), the second inequality of (7.119) can be obtained. �

Remark 7.21

(i) By setting p = 0, = m = 1, h(t) = t and g = I, [2, Theorem 3.1] is obtained

(ii) By setting = m = 1, h(t) = t and g = I, [64, Theorem 2.1] is obtained.

(iii) By setting w = p = 0, = m = 1, h(t) = t and g = I, [78, Theorem 4] is obtained.

Theorem 7.31 Let f ,g, h : [a,b]→R, 0 < a < b, Range(g),Range(h)⊂ [a,b] be such thatf is positive and f ∈ L1[a,b], g differentiable, strictly increasing and h nonnegative and

integrable. If f is harmonically (,h−m)-convex and f(

1g(x)

)= f

(1

1g(a) + 1

mg(b)−mg(x)

),

then for operators (2.23) and (2.24) we have:

f


)⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′, h ◦) (7.125)

≤(

h

(12

)+mh

(1− 1

2

))⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,(h◦)( f ◦))

≤(

mg(b)−g(a)m2g(a)g(b)

){m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)h(t)dt

+(

h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)h(1− t)dt

},

where w′ = w(


), (t) = 1


a ].

Proof. Multiplying (7.121) by t−1h

(mg(a)g(b)

tg(b)+(1−t) g(a)m

)E ,c,q,r , ,, (w(t); p), and then inte-

grating over [0,1] we get

f


)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)dt (7.126)

≤ h

(12

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)f

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)dt

+mh

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

tg(b)+ (1−t) g(a)m

)f

(mg(a)g(b)

tg(a)+m(1−t)g(b)

)dt.


By choosing g(x) = tg(b)+(1−t) g(a)m

mg(a)g(b) that is mg(a)g(b)tg(a)+m(1−t)g(b) = 1

1g(a) +

1mg(b)−mg(x)

in (7.126) and

using (2.23) and (2.24), and the condition f(

1g(x)

)= f

(1

1g(a) +

1mg(b)−mg(x)

), the first in-

equality of (7.125) can be obtained.For the second inequality of (7.125), multiplying (7.123) by

t−1E ,c,q,r , ,, (wt ; p)h

(mg(a)g(b)

tg(b)+(1−t) g(a)m

), then integrating over [0,1] we get

h

(12

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)f

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)dt

(7.127)

+mh

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)f

(mg(a)g(b)

tg(a)+m(1− t)g(b)

)dt

≤ m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)

× h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)h(t)dt +

(h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)

×∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h(t)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)h(1− t)dt.


mg(a)g(b) in (7.127) and using (2.23) and (2.24), and the condition

f(

1g(x)

)= f

(1

1g(a) + 1

mg(b)−mg(x)

), the second inequality of (7.125) can be obtained. �

Remark 7.22

(i) By setting p = 0, = m = 1, h(t) = t h(x) = 1 and g = I, [2, Theorem 3.1] isobtained.

(ii) By setting g = I, = m = 1, h(t) = t and h(x) = 1, [64, Theorem 2.1] is obtained.

(iii) By setting w = p = 0, h(x) = 1, = m = 1, h(t) = t and g = I, [78, Theorem 4] isobtained.

(iv) By setting w = p = 0, = 1, = m = 1, h(t) = t and g = I, [33, Theorem 8] isobtained.

(v) By setting w = p = 0, = 1, h(x) = 1, = m = 1, h(t) = t and g = I, [95, Theorem2.4] is obtained.


Theorem 7.32 Let f ,g : [a,b]→R, 0 < a < b, Range(g)⊂ [a,b] be such that f is positiveand f ∈ L1[a,b] and g differentiable and strictly increasing. If f is harmonically (,h−m)-convex on [a,b], then for operators (2.23) and (2.24) we have:

f


)⎛⎝Fg−1

(g(a)+mg(b)2m2g(a)g(b)

)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′,1) (7.128)

≤ h

(12

)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′, f ◦)

+m1−2h

(1− 1

2

)(F

g−1(

g(a)+mg(b)2mg(a)g(b)

)+

g−1(

1g(a)

),

)(2w′, f ◦)

≤(

mg(b)−g(a)2m2g(a)g(b)

) {m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h(( t

2

))dt +

(h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h(

1−( t

2

))dt

}, w′ = w

(mg(a)(b)

mg(b)−g(a)

),

where (t) = 1g(t) and t ∈ [ 1

b , 1a ].

Proof. By setting g(x) = mg(a)g(b)t2 g(b)+(1− t

2 ) g(a)m

and g(y) = mg(a)g(b)t2 g(a)+m(1− t

2 )g(b) in (7.120), we have

f


)≤ h

(12

)f

(mg(a)g(b)

t2g(b)+ (1− t

2) g(a)m

)(7.129)

+mh

(1− 1

2

)f

(mg(a)g(b)

t2g(a)+m(1− t

2)g(b)

).

Multiplying (7.129) by t−1E ,c,q,r , ,, (wt ; p) and then integrating over [0,1], we get

f


)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)dt (7.130)

≤ h

(12

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)dt

+mh

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

t2g(a)+m(1− t

2 )g(b)

)dt.

By setting g(x) =t2 g(b)+(1− t

2 ) g(a)m

mg(a)g(b) and g(y) =t2 g(a)+m(1− t

2 )g(b)mg(a)g(b) in (7.130) and using (2.23)

and (2.24), first inequality of (7.128) can be obtained. On the other hand from harmonically


(,h−m)-convexity of f we have

h

(12

)f

(mg(a)g(b)

t2g(b)+ (1− t

2) g(a)m

)+mh

(1− 1

2

)f

(mg(a)g(b)

t2g(a)+m(1− t

2 )g(b)

)(7.131)

≤ h

(12

)(h

(1−

( t2

))f (g(b))+mh

(( t2

))f

(g(a)m2

))

+mh

(1− 1

2

)(h

(1−

( t2

))f (g(a))+mh

(( t2

))f (g(b))

)

= m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)h

(( t2

))

+(

h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)h

(1−

( t2

)).

Multiplying (7.131) by t−1E ,c,q,r , ,, (wt ; p) and integrating over [0,1], we get

h

(12

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)dt (7.132)

+mh

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p) f

(mg(a)g(b)

t2g(a)+m(1− t

2 )g(b)

)dt

≤ m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)

×∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h(( t

2

))dt +

(h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)

×∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h(

1−( t

2

))dt.

By setting g(x) =t2 g(b)+(1− t

2 ) g(a)m

mg(a)g(b) and g(y) =t2 g(a)+m(1− t

2 )g(b)mg(a)g(b) in (7.132) and using (2.23)

and (2.24), second inequality of (7.128) can be obtained. �

Remark 7.23

(i) By setting p = 0, = m = 1, h(t) = t and g = I, [2, Theorem 3.3] is obtained.


Theorem 7.33 Let f ,g, h : [a,b] → R, 0 < a < mb, Range(g),Range(h) ⊂ [a,b] be suchthat f is positive and f ∈ L1[a,b], g differentiable, strictly increasing and h nonnegative

and integrable. If f is harmonically ( ,h-m)-convex and f(

1g(x)

)= f

(1

1g(a) +

1mg(b)−mg(x)

),

then for operators (2.23) and (2.24) we have:


f


)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′, h◦) (7.133)

≤(

h

(12

)+mh

(1− 1

2

))⎛⎝Fg−1


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′,(h ◦)( f ◦))

≤(

mg(b)−g(a)2m2g(a)g(b)

){m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)h

(( t2

))dt

+(

h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h

(mg(a)g(b)

t2g(b)+ (1− t

2) g(a)m

)h

(1−

( t2

))dt

},

where w′ = w(


), (t) = 1


a ].

Proof. Multiplying (7.129) by t−1h

(mg(a)g(b)

t2 g(b)+(1− t

2 ) g(a)m

)E ,c,q,r , ,, (wt ; p) and integrating

over [0,1] we get

f


)∫ 1

0t−1h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)E ,c,q,r , ,, (wt ; p)dt ≤ h

(12

)(7.134)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)f

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)dt +m

h

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)f

(mg(a)g(b)

t2g(a)+m(1− t

2 )g(b)

)dt.

By choosing g(x) = mg(a)g(b)t2 g(b)+(1− t

2 ) g(a)m

in (7.134) and using (2.23) and (2.24), and the condi-

tion f(

1g(x)

)= f

(1

1g(a) + 1

mg(b)−mg(x)

), then first inequality of (7.133) can be obtained.

On the other hand by harmonically ( ,h-m)-convexity of f , multiplying (7.131) by

t−1h

(mg(a)g(b)

t2 g(b)+(1− t

2 ) g(a)m

)E ,c,q,r , ,, (wt ; p) and integrating over [0,1], we get


h

(12

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)f

(mg(a)g(b)

t2g(b)+ (1− t

2) g(a)m

)dt

(7.135)

+mh

(1− 1

2

)∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)

× f

(mg(a)g(b)

t2g(a)+m(1− t

2)g(b)

)dt ≤ m

(h

(12

)f

(g(a)m2

)+mh

(1− 1

2

)f (g(b))

)

×∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)h

(( t2

))dt

+(

h

(12

)f (g(b))+mh

(1− 1

2

)f (g(a))

)

×∫ 1

0t−1E ,c,q,r

, ,, (wt ; p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)h

(1−

( t2

))dt.

By choosing g(x) = mg(a)g(b)t2 g(b)+(1− t

2 ) g(a)m

in (7.135) and using (2.23) and (2.24), and the condi-

tion f(

1g(x)

)= f

(1

1g(a) +

1mg(b)−mg(x)

)second inequality of (7.133) can be obtained. �

Remark 7.24

(i) By setting p = 0, = m = 1, h(t) = t and g = I, [2, Theorem 3.6] is obtained.


7.6.1 Results for Harmonically (h−m)-convex Functions

By setting = 1 in Theorem 7.30-Theorem 7.33, the results for harmonically (h−m)-convex functions are obtained as follows:

Theorem 7.34 Under the assumptions of Theorem 7.30, the following inequality holdsfor harmonically (h−m)-convex functions:

f


)⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,1) (7.136)

≤ h

(12

)⎧⎨⎩⎛⎝F

g−1(

1mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′, f ◦)+m1−2

(F

g−1(

1mg(b)

)+

g−1(

1g(a)

),

)(w′, f ◦)

⎫⎬⎭

≤ h

(12

)(mg(b)−g(a)m2g(a)g(b)

){m

(f

(g(a)m2

)+mf (g(b))

)∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h(t)dt +( f (g(b))+mf (g(a)))∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h(1− t)dt

}.



f


)⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′, h◦) (7.137)

≤ h

(12

)(1+m)

⎛⎝F

g−1(

1mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,(h◦)( f ◦))

≤ h

(12


){m

(f

(g(a)m2

)+mf (g(b))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)h(t)dt

+( f (g(b))+mf (g(a)))∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)h(1− t)dt

}.


f


)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′,1) ≤ h

(12

)(7.138)

⎧⎨⎩⎛⎝F

g−1(

g(a)+mg(b)2m2g(a)g(b)

)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′, f ◦)+m1−2

(F

g−1(


)+

g−1(

1g(a)

),

)(2w′, f ◦)

⎫⎬⎭

≤ h

(12

)(mg(b)−g(a)2m2g(a)g(b)

){m

(f

(g(a)m2

)+mf (g(b))

)∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h( t

2

)dt

+( f (g(b))+mf (g(a)))∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h(1− t

2

)dt

}.


f


)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′, h◦) (7.139)

≤ h

(12

)(1+m)

⎛⎝F

g−1(


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′,(h◦)( f ◦))

≤ h

(12


) {m

(f

(g(a)m2

)+mf (g(b))

)


∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h

(mg(a)g(b)

t2g(b)+ (1− t

2) g(a)m

)h( t

2

)dt +( f (g(b))+mf (g(a)))

∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h

(mg(a)g(b)

t2g(b)+ (1− t

2) g(a)m

)h(1− t

2

)dt

}.

7.6.2 Results for Harmonically (,m)-convex Functions

By setting h(t) = t in Theorem 7.30-Theorem 7.33, the results for harmonically (-m)-convex functions are obtained as follows:

Theorem 7.38 Under the assumptions of Theorem 7.30, the following inequality holdsfor harmonically (,m)-convex functions:

f


)⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,1) (7.140)

≤(

12

)⎧⎨⎩⎛⎝F

g−1(

1mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′, f ◦)+m1−2 (2 −1)

(F

g−1(

1mg(b)

)+

g−1(

1g(a)

),

)(w′, f ◦)

⎫⎬⎭

≤(

12


){m

(f

(g(a)m2

)+m(2 −1) f (g(b))

)

×⎛⎝F

g−1(

1mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,1)+ ( f (g(b))+m(2 −1) f (g(a)))

×⎛⎝⎛⎝F

g−1(

1mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,1)−

⎛⎝F

g−1(

1mg(a)

)−g−1

(1

m2g(b)

),+

⎞⎠(mw′,1)

⎞⎠⎫⎬⎭ .


f


)⎛⎝Fg−1

(1

mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′, h ◦)≤

(12

)(1+m) (7.141)

⎛⎝F

g−1(

1mg(a)

)−g−1

(1

m2g(b)

),

⎞⎠(mw′,(h◦)( f ◦)) ≤

(12


)

{m

(f

(g(a)m2

)+m(2 −1) f (g(b))

) ∫ 1

0t+−1E ,c,q,r

, ,, (w(t ); p)

h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)dt +( f (g(b))+m(2 −1) f (g(a)))

×∫ 1

0(t−1− t+−1)E ,c,q,r

, ,, (w(t); p)h

(mg(a)g(b)

tg(b)+ (1− t) g(a)m

)dt

}.



f


)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′,1) ≤

(12

)(7.142)

⎧⎨⎩⎛⎝F

g−1(


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′, f ◦)+m1−2

(F

g−1(


)+

g−1(

1g(a)

),

)(2w′, f ◦)

⎫⎬⎭

≤(

12


) {m

(f

(g(a)m2

)+m(2 −1) f (g(b))

)

×(

12

)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),+

⎞⎠((2m)w′,1)+ ( f (g(b))+m(2 −1) f (g(a)))

×⎛⎝⎛⎝F

g−1(


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′,1)−

(12

)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),+

⎞⎠((2m)w′,1)

⎞⎠⎫⎬⎭ .


f


)⎛⎝Fg−1


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′, h◦) (7.143)

≤(

12

)(1+m)

⎛⎝F

g−1(


)−

g−1(

1m2g(b)

),

⎞⎠((2m)w′,(h ◦)( f ◦))

≤(

12


){m

(f

(g(a)m2

)+m(2 −1) f (g(b))

)

×∫ 1

0t−1E ,c,q,r

, ,, (w(t); p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)( t2

)dt

+( f (g(b))+m(2 −1) f (g(a)))

×∫ 1

0t−1E ,c,q,r

, ,, (w(t ); p)h

(mg(a)g(b)

t2g(b)+ (1− t

2 ) g(a)m

)(1−

( t2

))dt

}.

Chapter8Error Bounds of Hadamardand Fejer-HadamardInequalities and Bounds ofFractional Integral OperatorsAssociated withMittag-Leffler Function

In this chapter fractional integral inequalities are given which provide bounds of variouskinds of fractional integral operators containing extended Mittag-Leffler functions. Alsoestimations of Hadamard and Fejer-Hadamard inequalities are given for different kinds ofconvex functions.

This chapter is based on our results from [35, 36, 55, 57, 54, 70, 101, 125, 132, 148,150].

133

134 8 BOUNDS OF HADAMARD AND FEJER-HADAMARD INEQUALITIES

8.1 Error Bounds Associated with FractionalIntegral Inequalities for Convex Functions

The following lemma is needed to prove the results of this section.

Lemma 8.1 Let f : [a,b] → R be a differentiable function such that f ′ ∈ L1[a,b] witha < b. Also let g : [a,b]→R be continuous on [a,b], then the following identity for extendedgeneralized fractional integral operators holds(∫ b

ag(s)E ,c,q,r

, , (ws ; p)ds

)[ f (a)+ f (b)] (8.1)

−∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

=∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)f ′(t)dt−

∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)f ′(t)dt.

Proof. On integrating by parts we have

∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)f ′(t)dt (8.2)

=(∫ b

ag(s)E ,c,q,r

, , (ws ; p)ds

)f (b)−

∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt,

and ∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)f ′(t)dt (8.3)

= −(∫ b

ag(s)E ,c,q,r

, , (ws ; p)ds

)f (a)+

∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt.

Subtracting (8.3) from (8.2) we get (8.1) which is required identity. �

In [15] Andric et al. proved the absolute convergence of the function E ,r,k,c ,, (t; p). If

we set

M :=

n=0

∣∣∣∣ p( +nk,c−)(c)nktn

( ,c−)(n+ )( )nr

∣∣∣∣ ,then |E ,r,k,c

,, (t; p)| ≤ M. We use this and the the identity (8.1) to prove the next results.

8.1 ERROR BOUNDS FOR CONVEX FUNCTIONS 135

Theorem 8.1 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R be a differentiable function such that f ′ ∈ L1[a,b] with a < b. Also letg : [a,b]→ R be a continuous function on [a,b]. If | f ′| is convex function on [a,b], then thefollowing inequality for extended generalized fractional integral operators holds∣∣∣∣

(∫ b

ag(s)E ,c,q,r

, , (ws ; p)ds

)[ f (a)+ f (b)] (8.4)

−∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ (b−a)+1‖g‖M

( +1)[| f ′(a)|+ | f ′(b)|]

for q < r+() and ‖ g ‖= supt∈[a,b]

|g(t)|.

Proof. From Lemma 8.1, we have∣∣∣∣(∫ b

ag(s)E ,,q

, ,r(ws ; p)ds

)[ f (a)+ f (b)] (8.5)

−∫ b

a

(∫ t

ag(s)E ,,q

, ,r(ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ b

a

(∫ b

tg(s)E ,,q

, ,r(ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤∫ b

a

∣∣∣∣∫ t

ag(s)E ,,q

, ,r(ws ; p)ds

∣∣∣∣| f ′(t)|dt +

∫ b

a

∣∣∣∣∫ b

tg(s)E ,,q

, ,r(ws ; p)ds

∣∣∣∣| f ′(t)|dt

Using absolute convergence of Mittag-Leffler function and ‖ g ‖= supt∈[a,b]

|g(t)|, we have

∣∣∣∣(∫ b

ag(s)E ,c,q,r

, , (ws ; p)ds

)[ f (a)+ f (b)] (8.6)

−∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ ‖g‖M

[∫ b

a(t−a) | f ′(t)|dt +

∫ b

a(b− t) | f ′(t)|dt

].

As | f ′| is convex function, for t ∈ [a,b] we have

| f ′(t)| ≤ b− tb−a

| f ′(a)|+ t −ab−a

| f ′(b)|. (8.7)


Using (8.7) in (8.6), we have∣∣∣∣(∫ b

ag(s)E ,c,q,r

, , (ws ; p)ds

)[ f (a)+ f (b)] (8.8)

−∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ ‖g‖M

[∫ b

a(t−a)

(b− tb−a

| f ′(a)|+ t−ab−a

| f ′(b)|)

dt

+∫ b

a(b− t)

(b− tb−a

| f ′(a)|+ t−ab−a

| f ′(b)|)

dt

].

After simple calculation of the above inequality, (8.4) can be obtained. �



(ii) If w = p = 0, then we get [137, Theorem 6].

(iii) If g(s) = 1 along with w = p = 0, then we get [137, Corollary 2].

Corollary 8.1 In Theorem 8.1, for w = 0, = and g(s) = 1 we have the followinginequality for Riemann-Liouville fractional integral operator∣∣∣∣ f (a)+ f (b)

2− (+1)

2(b−a)[Ja+ f (b)+ Jb− f (a)]

∣∣∣∣≤ b−a2(+1)

[| f ′(a)|+ | f ′(b)|]

> 0.

Theorem 8.2 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R be a differentiable function such that f ′ ∈ L1[a,b] with a < b. Also letg : [a,b] → R be a continuous function on [a,b]. If | f ′|q is convex function on [a,b], thenfor q > 0 the following inequality for extended generalized fractional integral operatorsholds ∣∣∣∣

(∫ b

ag(s)E ,c,q,r

, , (ws ; p)ds

)[ f (a)+ f (b)] (8.9)

−∫ b

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ b

a

(∫ b

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ 2(b−a)+1‖g‖M

( p+1)1q

[ | f ′(a)|q + | f ′(b)|q2

] 1q


for q < r+() and ‖ g ‖= supt∈[a,b]

|g(t)| and 1p + 1

q = 1.

Proof. The proof is on the same lines as the proof of Theorem 8.1. �




(iii) If w = p = 0, = 1, then we get [137, Corollary 3].

(iv) If g(s) = 1 and w = p = 0, then we get [42, Theorem 2.3].

Corollary 8.2 In Theorem 8.2, if we take w = 0 and g(s) = 1, then we have the followinginequality of Riemann-Liouville integral for fractional integral operator

∣∣∣∣ f (a)+ f (b)2

− ( +1)2(b−a)

[Ja+ f (b)+ Jb− f (a)]∣∣∣∣≤ b−a

( p+1)1p

[ | f ′(a)|q + | f ′(b)|q2

] 1q

> 0.

Lemma 8.2 Let f ,g : [a,b] → R, 0 < a < b be the functions such that f be positive andf ∈ L1[a,b] and g be a differentiable and strictly increasing. If the function g is symmetricabout a+b

2 , then we have(g ,r,k,c

,, ,w,a+ f ◦ g)

(b; p) =(

g ,r,k,c ,, ,w,b− f ◦ g

)(a; p) (8.10)

=12

[(g ,r,k,c

,, ,w,a+ f ◦ g)

(b; p)+(

g ,r,k,c ,, ,w,b− f ◦ g

)(a; p)

].

Proof. Since g is symmetric about a+b2 , by Definition 2.4 of extended generalized frac-

tional integral operator, we have

(g

,r,k,c ,, ,w,a+ f ◦ g

)(b; p) =

∫ b

a(g(b)−g(t))−1 E ,r,k,c

,, (w(g(b)−g(t)) ; p) f (g(t))d(g(t)).

(8.11)

If we replace t by a+b− t in (8.11), then we get

(g

,r,k,c ,, ,w,a+ f ◦ g

)(b; p) =

∫ b

a(g(t)−g(a))−1 E ,r,q,c

,, (w(g(t)−g(a)) ; p) f (g(t))d(g(t)).

This implies (g

,r,k,c ,, ,w,a+ f ◦ g

)(b; p) =

(g

,r,k,c ,, ,w,b− f ◦ g

)(a; p). (8.12)

By adding equations (8.11) and (8.12), we get (8.10). �


Lemma 8.3 Let f ,g,h : [a,b]→ R, 0 < a < b be the functions such that f be positive andf ◦ g ∈ L1[a,b], where g be a differentiable and strictly increasing and h be continuous.For f ′ ◦ g ∈ L1[a,b], the following equality for extended generalized fractional integraloperators (2.23) and (2.24) holds:(

f (g(a))+ f (g(b))2

)[(g

,r,k,c ,, ,w,a+h

)(b; p)+

(g

,r,k,c ,, ,w,b−h

)(a; p)

](8.13)

−[(

g ,r,k,c ,, ,w,a+h( f ◦ g)

)(b; p)+

(g

,r,k,c ,, ,w,b−h( f ◦ g)

)(a; p)

]=∫ b

a

[∫ t

a(g(b)−g(s))−1E ,r,k,c

,, (w(g(b)−g(s)) ; p)h(s)d(g(s))

−∫ b

t(g(s)−g(a))−1E ,r,k,c

,, (w(g(s)−g(a)) ; p)h(s)d(g(s))]

f ′(g(t))d(g(t)).

Proof. To prove this lemma, we take its right hand side. On integrating by parts and aftersimplification, we have

∫ b

a

[∫ t


,, (w(g(b)−g(s)) ; p)h(s)d(g(s))]

f ′(g(t))d(g(t))

= f (g(b))(∫ b


,, (w(g(b)−g(s)) ; p)h(s)d(g(s)))

−∫ b

a

((g(b)−g(t))−1E ,r,k,c

,, (w(g(b)−g(t)) ; p))

h(t) f (g(t))d(g(t))

= f (g(b))(

g ,r,k,c ,, ,w,a+h

)(b; p)−

(g

,r,k,c ,, ,w,a+h( f ◦ g)

)(b; p).

By using Lemma 8.2, we have

∫ b

a

[∫ t


,, (w(g(b)−g(s)) ; p)h(s)d(g(s))]

f ′(g(t))d(g(t)) (8.14)

=f (g(b))

2

[(g

,r,k,c ,, ,w,a+h

)(b; p)+

(g

,r,k,c ,, ,w,b−h

)(a; p)

]−(

g ,r,k,c ,, ,w,a+h( f ◦ g)

)(b; p).

Similarly

∫ b

a

[−∫ b


,, (w(g(s)−g(a)) ; p)h(s)d(g(s))]

f ′(g(t))d(g(t))

(8.15)

=f (g(a))

2

[(g ,r,k,c

,, ,w,a+h)

(b; p)+(

g ,r,k,c ,, ,w,b−h

)(a; p)

]−(

g ,r,k,c ,, ,w,b−h( f ◦ g)

)(a; p).

Summing (8.14) and (8.15), we get (8.13). �

Theorem 8.3 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,g,h : [a,b]→R, 0 < a < b be the functions such that f be positive and ( f ◦g)′ ∈ L1[a,b],where g be a differentiable and strictly increasing and h be continuous and symmetric


about a+b2 . Also let f ◦ g be symmetric about a+b

2 and if |( f ◦ g)′| is convex. Then fork < r+(), the following inequality holds:

∣∣∣∣(

f (g(a))+ f (g(b))2

)[(g

,r,k,c ,, ,w,a+h

)(b; p)+

(g

,r,k,c ,, ,w,b−h

)(a; p)

]−[(

g ,r,k,c ,, ,w,a+( f ◦ g)h

)(b; p)+

(g ,r,k,c

,, ,w,b−( f ◦ g)h)

(a; p)]∣∣∣

≤ ‖ h ‖ M(g(b)−g(a))+1

( +1)(1−)

[| f ′(g(a))+ f ′(g(b))|] ,where ‖ h ‖= sup

t∈[a,b]|h(t)| and

=1

+2

⎡⎣(

g( a+b2 )−g(a)

g(b)−g(a)

)+2

+

(g(b)−g( a+b

2 )g(b)−g(a)

)+2⎤⎦

− +1 +2

⎡⎣(

g( a+b2 )−g(a)

g(b)−g(a)

)+2

+

(g(b)−g( a+b

2 )g(b)−g(a)

)+2⎤⎦

−(

g( a+b2 )−g(a)

g(b)−g(a)

)+1(g(b)−g( a+b

2 )g(b)−g(a)

)+

(g( a+b

2 )−g(a)g(b)−g(a)

)(g(b)−g( a+b

2 )g(b)−g(a)

)+1

.

Proof. By using Lemma 8.3, we have

∣∣∣∣(

f (g(a))+ f (g(b))2

)[(g

,r,k,c ,, ,w,a+h

)(b; p)+

(g

,r,k,c ,, ,w,b−h

)(a; p)

](8.16)

−[(

g ,r,k,c ,, ,w,a+( f ◦ g)h

)(b; p)+

(g

,r,k,c ,, ,w,b−( f ◦ g)h

)(a; p)

]∣∣∣≤∫ b

a

∣∣∣∣[∫ t


,, (w(g(b)−g(s)) ; p)h(s)d(g(s))

−∫ b


,, (w(g(s)−g(a)) ; p)h(s)d(g(s))]∣∣∣∣ | f ′(g(t))|d(g(t)).

Using the convexity of | f ′(g)| on [a,b], we have

| f ′(g(t))| ≤ g(b)−g(t)g(b)−g(a)

| f ′(g(a))|+ g(t)−g(a)g(b)−g(a)

| f ′(g(b))|, t ∈ [a,b]. (8.17)

The symmetry of h implies h(t) = h(a + b− t), h(s) = h(a + b− s) and replacing t bya+b− t, s by a+b− s in second integral, we get


∣∣∣∣∫ t


,, (w(g(b)−g(s)) ; p)h(s)d(g(s))

−∫ b

t(g(s)−g(a))−1E ,c,q,r

, , (w(g(s)−g(a)) ; p)h(s)d(g(s))∣∣∣∣

=∣∣∣∣−

∫ a

t(g(b)−g(s))−1E ,c,q,r

, , (w(g(b)−g(s)) ; p)h(s)d(g(s))

−∫ a+b−t

a(g(b)−g(s))−1E ,c,q,r

, , (w(g(b)−g(s)) ; p)h(s)d(g(s))∣∣∣∣

=∣∣∣∣∫ a+b−t

t(g(b)−g(s))−1E ,c,q,r

, , (w(g(b)−g(s)) ; p)h(s)d(g(s))∣∣∣∣

≤

⎧⎪⎪⎨⎪⎪⎩

∫ a+b−tt |(g(b)−g(s))−1E ,c,q,r

, , (w(g(b)−g(s)) ; p)h(s)|d(g(s)), t ∈ [a, a+b2 ]

∫ ta+b−t |(g(b)−g(s))−1E ,c,q,r

, , (w(g(b)−g(s)) ; p)h(s)|d(g(s)), t ∈ [ a+b2 ,b].

(8.18)By (8.16), (8.17), (8.18) and absolute convergence of Mittag-Leffler function, we have

∣∣∣∣(

f (g(a))+ f (g(b))2

)[(g

,r,k,c ,, ,w,a+h

)(b; p)+

(g

,r,k,c ,, ,w,b−h

)(a; p)

](8.19)

−[(

g ,r,k,c ,, ,w,a+( f ◦ g)h

)(b; p)+

(g

,r,k,c ,, ,w,b−( f ◦ g)h

)(a; p)

]∣∣∣≤∫ a+b

2

a

(∫ a+b−t

a|(g(b)−g(s))−1E ,c,q,r

, , (w(g(b)−g(s)) ; p)h(s)|d(g(s)))

×(

g(b)−g(t)g(b)−g(a)

| f ′(g(a))|+ g(t)−g(a)g(b)−g(a)

| f ′(g(b))|)

d(g(t))

+∫ b

a+b2

(∫ t

a+b−t|(g(b)−g(s))−1E ,c,q,r

, , (w(g(b)−g(s)) ; p)h(s)|d(g(s)))

×(


| f ′(g(a))|+ g(t)−g(a)g(b)−g(a)

| f ′(g(b))|)

d(g(t)).

≤ ‖ h ‖ M(g(b)−g(a))

[∫ a+b2

a((g(b)−g(t)) − (g(t)−g(a))(g(b)−g(t))| f ′(g(a))|)d(g(t))

+∫ a+b

2

a((g(b)−g(t)) − (g(t)−g(a))(g(t)−g(a))| f ′(g(b))|)d(g(t))

+∫ b

a+b2

((g(t)−g(a)) − (g(b)−g(t))(g(b)−g(t))| f ′(g(a))|)d(g(t))

+∫ b

a+b2

((g(t)−g(a)) − (g(b)−g(t))(g(t)−g(a))| f ′(g(b))|)d(g(t))].


After solving the terms of above inequality, we have the following values∫ a+b2

a((g(b)−g(t)) − (g(t)−g(a))) (g(b)−g(t))d(g(t))

=∫ b

a+b2

((g(t)−g(a)) − (g(b)−g(t))) (g(t)−g(a))d(g(t))

=(g(b)−g(a))+2

+2−(g(b)−g( a+b

2 ))+2

+2

−(g( a+b

2 )−g(a))+1

+1

(g(b)−g(

a+b2

))−(g( a+b

2 )−g(a))+2

( +1)(+2)

and ∫ a+b2

a((g(b)−g(t))− (g(t)−g(a)))(g(t)−g(a))d(g(t))

=∫ b

a+b2

((g(t)−g(a))− (g(b)−g(t)))(g(b)−g(t))d(g(t))

= −(g( a+b

2 )−g(a))+1

+1

(g(b)−g(

a+b2

))

+(g(b)−g(a))+2

( +1)(+2)−(g( a+b

2 )−g(a))+2

( +1)(+2)−(g(b)−g( a+b

2 ))+2

+2.

Using the above calculations of integrals in (8.19), we get the required inequality. �

Remark 8.3

(i) In Theorem 8.3 if we put g = I, we get [33, Theorem 2.3],

(ii) In Theorem 8.3 if we put p = 0 and g = I, we get [1, Theorem 2.3].

(iii) In Theorem 8.3 if we put w = p = 0 and g = I, we get [94, Theorem 2.36].

Theorem 8.4 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,g,h : [a,b]→R, 0 < a < b be the functions such that f be positive and ( f ◦g)′ ∈ L1[a,b],where g be a differentiable and strictly increasing and h be continuous and symmetricabout a+b

2 . Also let f ◦ g be symmetric about a+b2 and if |( f ◦ g)′|q, q ≥ 1 is convex on

[a,b]. Then for k < r+(), the following inequality holds:∣∣∣∣(

f (g(a))+ f (g(b))2

)[(g

,r,k,c ,, ,w,a+h

)(b; p)+

(g

,r,k,c ,, ,w,b−h

)(a; p)

]−[(

g ,r,k,c ,, ,w,a+( f ◦ g)h

)(b; p)+

(g ,r,k,c

,, ,w,b−( f ◦ g)h)

(a; p)]∣∣∣

≤2 ‖ h ‖ M(g(b)−g(a))+1

( +1)

[(1−)1−

1q (1−)

1q

]( | f ′(g(a))|q + | f ′(g(b))|q2

) 1q

,

(8.20)

where ‖ h ‖= supt∈[a,b]

|h(t)|, =(

g(b)−g( a+b2 )

g(b)−g(a)

)+1

+(

g( a+b2 )−g(a)

g(b)−g(a)

)+1

and


= 1+2

[(g( a+b

2 )−g(a)g(b)−g(a)

)+2

+(

g(b)−g( a+b2 )

g(b)−g(a)

)+2]

− +1+2

[(g( a+b

2 )−g(a)g(b)−g(a)

)+2

+(

g(b)−g( a+b2 )

g(b)−g(a)

)+2]

−(

g( a+b2 )−g(a)

g(b)−g(a)

)+1(g(b)−g( a+b

2 )g(b)−g(a)

)+(

g( a+b2 )−g(a)

g(b)−g(a)

)(g(b)−g( a+b

2 )g(b)−g(a)

)+1

.

Proof. Using Lemma 8.3, Holder inequality, (8.18) and convexity of | f ′(g)|q respectively,we get

∣∣∣∣(

f (g(a))+ f (g(b))2

)[(g ,r,k,c

,, ,w,a+h)

(b; p)+(

g ,r,k,c ,, ,w,b−h

)(a; p)

](8.21)

−[(

g ,r,k,c ,, ,w,a+( f ◦ g)h

)(b; p)+

(g

,r,k,c ,, ,w,b−( f ◦ g)h

)(a; p)

]∣∣∣≤[∣∣∣∣∫ b

a

(∫ a+b−t

t(g(b)−g(s))−1E ,r,k,c

,, (w(g(b)−g(s)) ))

h(s)d(g(s))∣∣∣∣d(g(t))

]1− 1q

[∣∣∣∣∫ b

a

(∫ a+b−t

t(g(b)−g(s))−1E ,r,k,c

,, (w(g(b)−g(s)) ))

h(s)d(g(s))∣∣∣∣ | f ′(g(t))|qd(g(t))

] 1q

.

Since | f ′(g)|q is convex on [a,b], we have

| f ′(g(t))|q ≤ g(b)−g(t)g(b)−g(a)

| f ′(g(a))|q +g(t)−g(a)g(b)−g(a)

| f ′(g(b))|q. (8.22)

Using ‖ h ‖= supt∈[a,b]

|h(t)|, and absolute convergence of Mittag-Leffler function, inequality

(8.21) becomes

∣∣∣∣ f (g(a))+ f (g(b))2

[(g

,r,k,c ,, ,w,a+h

)(b; p)+

(g

,r,k,c ,, ,w,b−h

)(a; p)

]−[(

g ,r,k,c ,, ,w,a+( f ◦ g)h

)(b; p)+

(g

,r,k,c ,, ,w,b−( f ◦ g)h

)(a; p)

]∣∣∣≤ ‖ h ‖ M

[∫ a+b2

a{(g(b)−g(t)) − (g(t)−g(b))}d(g(t))

+∫ b

a+b2

{(g(t)−g(a)) − (g(b)−g(t))}d(g(t))]1− 1

q

×[∫ a+b

2

a{(g(b)−g(t)) − (g(t)−g(b))}

×(


| f ′(g(a))|q +g(t)−g(a)g(b)−g(a)

| f ′(g(b))|q)

d(g(t))

8.2 ERROR BOUNDS FOR m-CONVEX FUNCTIONS 143

+∫ b

a+b2

{(g(t)−g(a)) − (g(b)−g(t))}

×(


| f ′(g(a))|q +g(t)−g(a)g(b)−g(a)

| f ′(g(b))|q)

d(g(t))] 1

q

.

After integrating and simplifying above inequality, we get (8.20). �

Remark 8.4

(i) In Theorem 8.4 if we put g = I, we get [33, Theorem 2.5].

(ii) In Theorem 8.4 if we put p = 0 and g = I, we get [1, Theorem 2.6].

(iii) In Theorem 8.4 if we put w = p = 0 and g = I, we get [94, Theorem 2.8].

8.2 Error Bounds Associated with FractionalIntegral Inequalities for m-convex Functions

The following lemma is needed to prove results of this section.

Lemma 8.4 Let f : [a,mb]→ R be a differentiable function such that f ′ ∈ L1[a,mb] with0≤ a < mb. Also let g : [a,mb]→R be a continuous function on [a,mb], then the followingidentity for extended generalized fractional integral operators holds(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)[ f (a)+ f (mb)] (8.23)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

=∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)f ′(t)dt−

∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)f ′(t)dt.

Proof. Proof is similar to the proof of Lemma 8.1. �

Theorem 8.5 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,mb] → R be a differentiable function such that f ′ ∈ L1[a,mb] with 0 ≤ a < mb.Also let g : [a,mb]→ R be a continuous function on [a,mb]. If | f ′| is m-convex function on[a,mb], then the following inequality for extended generalized fractional integral operatorsholds


∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.24)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ (mb−a)+1‖g‖M

( +1)(| f ′(a)|+m| f ′(b)|)

for q < r+() and ‖ g ‖= supt∈[a,mb]

|g(t)|.

Proof. From Lemma 2.22, we have

∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.25)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤∫ mb

a

∣∣∣∣∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

∣∣∣∣| f ′(t)|dt +

∫ mb

a

∣∣∣∣∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

∣∣∣∣| f ′(t)|dt.


|g(t)|, we have

∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.26)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ ‖g‖M

(∫ mb

a(t−a) | f ′(t)|dt +

∫ mb

a(mb− t) | f ′(t)|dt

).

As | f ′| is convex function, for t ∈ [a,mb] we have

| f ′(t)| ≤ mb− tmb−a

| f ′(a)|+mt−a

mb−a| f ′(b)|. (8.27)


Using (8.27) in (8.26), we have∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.28)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ ‖g‖M

(∫ mb

a(t −a)

(mb− tmb−a

| f ′(a)|+mt −a

mb−a| f ′(b)|

)dt

+∫ mb

a(mb− t)

(mb− tmb−a

| f ′(a)|+mt−a

mb−a| f ′(b)|

)dt

).

After simple calculation of above inequality we get (8.24) which is required. �

Remark 8.5

(i) If p = 0 in Theorem 8.5, then we get [53, Theorem 3.2]. For m = 1, Theorem 8.1 isobtained. Also for m = 1.


(iii) If g(s) = 1 along with w = p = 0 and = , then we get [137, Corollary 2].

Next we give another fractional integral inequality.

Theorem 8.6 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,mb] → R be a differentiable function such that f ∈ L1[a,mb] with 0 ≤ a < mb. Alsolet g : [a,mb] → R be a continuous function on [a,mb]. If | f ′|q is convex function on[a,mb], then for q > 0 the following inequality for extended generalized fractional integraloperators holds∣∣∣∣

(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.29)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ 2(mb−a)+1‖g‖M

( p+1)1q

( | f ′(a)|q +m| f ′(b)|q2

) 1q

for q < r+() and ‖ g ‖= supt∈[a,mb]

|g(t)| and 1p + 1

q = 1.


Proof. From Lemma 2.22, we have∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.30)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤∫ mb

a

∣∣∣∣∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

∣∣∣∣| f ′(t)|dt

+∫ mb

a

∣∣∣∣∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

∣∣∣∣| f ′(t)|dt.

By using Holder inequality we have∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.31)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤(∫ mb

a

∣∣∣∣∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

∣∣∣∣ p

dt

) 1p(∫ mb

a| f ′(t)|qdt

) 1q

+(∫ mb

a

∣∣∣∣∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

∣∣∣∣ p

dt

) 1p(∫ mb

a| f ′(t)|qdt

) 1q

.


|g(t)|, we have

∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.32)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ ‖g‖M

((∫ mb

a|t−a| pdt

) 1p

+(∫ mb

a|mb− t| pdt

) 1p)(∫ mb

a| f ′(t)|qdt

) 1q

.

As | f ′(t)|q is m-convex, for t ∈ [a,mb] we have

| f ′(t)|q ≤ mb− tmb−a

| f ′(a)|q +mt−a

mb−a| f ′(b)|q. (8.33)


Using (8.33) in (8.32), we have∣∣∣∣(∫ mb

ag(s)E ,c,q,r

, , (ws ; p)ds

)( f (a)+ f (mb)) (8.34)

−∫ mb

a

(∫ t

ag(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

−∫ mb

a

(∫ mb

tg(s)E ,c,q,r

, , (ws ; p)ds

)−1

g(t)E ,c,q,r , , (wt ; p) f (t)dt

∣∣∣∣∣≤ ‖g‖M

((∫ mb

a|t−a| pdt

) 1p

+(∫ mb

a|mb− t| pdt

) 1p)

×(∫ mb

a

mb− tmb−a

| f ′(a)|q +mt −a

mb−a| f ′(b)|q

) 1q

.

After simple calculation of above inequality, the inequality (8.29) which is obtained. �

Remark 8.6

(i) If p = 0 in Theorem 8.6, then we get [53, Theorem 3.6]. For m = 1, we get Theorem8.2. Also for m = 1.


(iii) If g(s) = 1 and w = p = 0, then we get [42, Theorem 2.3].

(iv) If w = p = 0 and = 1, then we get [42, Corollary 3].

To find error estimates first we prove the following two lemmas.

Lemma 8.5 Let f ,g : [a,mb] → R, 0 < a < mb, Range (g) ⊂ [a,mb] be the functionssuch that f be positive and f ◦ g ∈ L1[a,mb] and g be a differentiable and strictly increas-ing. Also if f (g(x)) = f (g(a)+ g(mb)− g(x)), then the following equality for fractionaloperators (2.23) and (2.24) holds:(

g ,r,k,c ,, ,w,a+ f ◦ g

)(mb; p) =

(g

,r,k,c ,, ,w,mb− f ◦ g

)(a; p)

=

(g

,r,k,c ,, ,w,a+ f ◦ g

)(mb; p)+

(g

,r,k,c ,, ,w,mb− f ◦ g

)(a; p)

2. (8.35)

Proof. By the definition of generalized fractional integral operator containing extendedMittag-Leffler function by a monotone function, we have(

g ,r,k,c ,, ,w,a+ f ◦ g

)(mb; p) (8.36)

=∫ mb

a(g(mb)−g(x))−1E ,r,k,c

,, (w(g(mb)−g(x)) ; p) f (g(x))d(g(x)),


Replacing g(x) by g(a)+ g(mb)− g(x) in (8.36) and using f (g(x)) = f (g(a)+ g(mb)−g(x)), we have

(g

,r,k,c ,, ,w,a+ f ◦ g

)(mb; p) =

∫ mb

a(g(x)−g(a))−1E ,r,k,c

,, (w(g(x)−g(a)) ; p) f (g(x))d(g(x)).

This implies (g

,r,k,c ,, ,w,a+ f ◦ g

)(mb; p) =

(g

,r,k,c ,, ,w,mb− f ◦ g

)(a; p). (8.37)

By adding (8.36) and (8.53), we get (8.35). �

Lemma 8.6 Let f ,g,h : [a,mb] → R, 0 < a < mb, Range (g), Range (h) ⊂ [a,mb] bethe functions such that f be positive and f ◦ g, h ◦ g ∈ L1[a,mb], g be a differentiableand strictly increasing and h be nonnegative and continuous. If f ′ ◦ g ∈ L1[a,mb] andh(g(t)) = h(g(a)+g(mb)−g(t)), then the following equality for the generalized fractionalintegral operators (2.23) and (2.24) holds:

f (g(a))+ f (g(mb))2

[(g ,r,k,c

,, ,w,a+h ◦ g)

(mb; p)+(

g ,r,k,c ,, ,w,mb−h ◦ g

)(a; p)

](8.38)

−[(

g ,r,k,c ,, ,w,a+( f ◦ g)(h ◦ g)

)(mb; p)+

(g

,r,k,c ,, ,w,mb−( f ◦ g)(h ◦ g)

)(a; p)

]=∫ mb

a

[∫ t

a(g(mb)−g(s))−1E ,r,k,c

,, (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))

−∫ mb


,, (w(g(s)−g(a)) ; p)h(g(s))d(g(s))]

f ′(g(t))d(g(t)).

Proof. To prove the lemma, we have

∫ mb

a

[∫ t


,, (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))]

f ′(g(t))d(g(t))

= f (g(mb))∫ mb


,, (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))

−∫ mb

a

((g(mb)−g(t))−1E ,r,k,c

,, (w(g(mb)−g(t)) ; p))

f (g(t))h(g(t))d(g(t))

= f (g(mb))(

g ,r,k,c ,, ,w,a+h ◦ g

)(mb; p)−

(g

,r,k,c ,, ,w,a+( f ◦ g)(h ◦ g)

)(mb; p).

By using Lemma 8.5, we have

∫ mb

a

[∫ t


,, (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))]

f ′(g(t))d(g(t))

=f (g(mb))

2

[(g

,r,k,c ,, ,w,a+h ◦ g

)(mb; p)+

(g

,r,k,c ,, ,w,mb−h ◦ g

)(a; p)

](

g ,r,k,c ,, ,w,a+( f ◦ g)(h ◦ g)

)(mb; p).


In the same way we have∫ mb

a

[−∫ mb


,, (w(g(s)−g(a)) ; p)h(g(s))d(g(s))]

f ′(g(t))d(g(t))

=f (g(a))

2

[(g

,r,k,c ,, ,w,a+h ◦ g

)(mb; p)+

(g

,r,k,c ,, ,w,mb−h ◦ g

)(a; p)

](

g ,r,k,c ,, ,w,mb−( f ◦ g)(h ◦ g)

)(a; p).

By adding (8.39) and (8.39), we get (8.38). �

By using Lemma 8.6, we prove the following theorem.

Theorem 8.7 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,g,h : [a,mb]→R, 0 < a < mb, Range (g), Range (h) ⊂ [a,mb] be the functions such thatf be positive and ( f ◦g)′ ∈ L1[a,mb], where g be a differentiable and strictly increasing andh be nonnegative and continuous. Also let h(g(t)) = h(g(a)+g(mb)−g(t)) and |( f ◦g)′| ism-convex on [a,b]. Then for k < r +(), the following inequality for fractional integraloperators (2.23) and (2.24) holds:∣∣∣∣ f (g(a))+ f (g(mb))

2

[(g

,r,k,c ,, ,w,a+h ◦ g

)(mb; p)+

(g

,r,k,c ,, ,w,mb−h ◦ g

)(a; p)

](8.39)

−[(

g ,r,k,c ,, ,w,a+( f ◦ g)(h ◦ g)

)(mb; p)+

(g

,r,k,c ,, ,w,mb−( f ◦ g)(h ◦ g)

)(a; p)

]∣∣∣≤ ‖ h ‖ M(g(mb)−g(a))+1

( +1)(1−)

[| f ′(g(a))+mf ′(g(b))|] ,where ‖ h ‖= sup

t∈[a,mb]|h(t)| and

= 1+2

[{(g( a+mb

2 )−g(a)g(mb)−g(a)

)+2}

+

{(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+2}]

− +1+2

[{(g( a+mb

2 )−g(a)g(mb)−g(a)

)+2}

+

{(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+2}]

−(

g( a+mb2 )−g(a)

g(mb)−g(a)

)+1(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+(

g( a+mb2 )−g(a)

g(mb)−g(a)

)(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+1

.

Proof. Using Lemma 8.6, we have∣∣∣∣(

f (g(a))+ f (g(mb))2

)[(g

,r,k,c ,, ,w,a+h ◦ g

)(mb; p)+

(g

,r,k,c ,, ,w,mb−h ◦ g

)(a; p)

]−[(

g ,r,k,c ,, ,w,a+( f ◦ g)(h ◦ g)

)(mb; p)+

(g

,r,k,c ,, ,w,mb−( f ◦ g)(h ◦ g)

)(a; p)

]∣∣∣≤∫ mb

a

∣∣∣∣[∫ t


,, (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))

−∫ mb


,, (w(g(s)−g(a)) ; p)h(g(s))d(g(s))]∣∣∣∣ | f ′(g(t))|d(g(t)).

(8.40)


Using the m-convexity of |( f ◦ g)′| on [a,b], we have

| f ′(g(t))| ≤ g(mb)−g(t)g(mb)−g(a)

| f ′(g(a))|+mg(t)−g(a)

g(mb)−g(a)| f ′(g(b))|, t ∈ [a,b]. (8.41)

If we replace g(s) by g(a)+ g(mb)− g(s) and using h(g(s)) = h(g(a)+ g(mb)− g(s)),t ′ = g−1(g(a)+g(mb)−g(t)), in second integral in the followings, we get∣∣∣∣

∫ t


,, (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))

−∫ mb

t(g(s)−g(a))−1E ,c,q,r

, , (w(g(s)−g(a)) ; p)h(g(s))d(g(s))∣∣∣∣

=∣∣∣∣−

∫ a

t(g(mb)−g(s))−1E ,c,q,r

, , (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))

−∫ t′

a(g(mb)−g(s))−1E ,c,q,r

, , (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))∣∣∣∣

=∣∣∣∣∫ t′

t(g(mb)−g(s))−1E ,c,q,r

, , (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))∣∣∣∣

≤

⎧⎪⎨⎪⎩∫ t′t |(g(mb)−g(s))−1E ,c,q,r

, , (w(g(mb)−g(s)) ; p)h(g(s))|d(g(s)), t ∈ [a, a+mb2 ]

∫ tt′ |(g(mb)−g(s))−1E ,c,q,r

, , (w(g(mb)−g(s)) ; p)h(g(s))|d(g(s)), t ∈ [ a+mb2 ,mb].

(8.42)By (8.40), (8.41), (8.42) and using absolute convergence of extended Mittag-Leffler func-tion, we have∣∣∣∣ f (g(a))+ f (g(mb))

2

((g

,r,k,c ,, ,w,a+h ◦ g

)(mb; p)+

(g

,r,k,c ,, ,w,mb−h ◦ g

)(a; p)

)−[(

g ,r,k,c ,, ,w,a+( f ◦ g)(h ◦ g)

)(mb; p)+

(g

,r,k,c ,, ,w,mb−( f ◦ g)(h ◦ g)

)(a; p)

]∣∣∣≤∫ a+mb

2

a

(∫ a+mb−t

a|(g(mb)−g(s))−1E ,r,k,c

,, (w(g(mb)−g(s)) ; p)h(g(s))|d(g(s)))

×(

g(mb)−g(t)g(mb)−g(a)

| f ′(g(a))|+mg(t)−g(a)

g(mb)−g(a)| f ′(g(b))|

)d(g(t))

+∫ mb

a+mb2

(∫ t

a+mb−t|(g(mb)−g(s))−1E ,r,k,c

,, (w(g(mb)−g(s)) ; p)h(g(s))|d(g(s)))

×(


| f ′(g(a))|+mg(t)−g(a)

g(mb)−g(a)| f ′(g(b))|

)d(g(t)).

≤ ‖ h ‖ M(g(mb)−g(a))

[∫ a+mb2

a((g(mb)−g(t))−(g(t)−g(a)))(g(mb)−g(t))| f ′(g(a))|d(g(t))

+m∫ a+mb

2

a((g(mb)−g(t)) − (g(t)−g(a)))m(g(t)−g(a))| f ′(g(b))|d(g(t))


+∫ mb

a+mb2

((g(t)−g(a)) − (g(mb)−g(t)))(g(mb)−g(t))| f ′(g(a))|d(g(t))

+m∫ mb

a+mb2

((g(t)−g(a)) − (g(mb)−g(t)))m(g(t)−g(a))| f ′(g(b))|d(g(t))]. (8.43)

After some calculations, we get

∫ a+mb2

a((g(mb)−g(t)) − (g(t)−g(a)))(g(mb)−g(t))d(g(t))

=∫ mb

a+mb2

((g(t)−g(a)) − (g(mb)−g(t)))(g(t)−g(a))d(g(t))

=(g(mb)−g(a))+2

+2−(g(mb)−g( a+mb

2 ))+2

+2−(g( a+mb

2 )−g(a))+1

+1(g(mb)−g(

a+mb2

))−(g( a+mb

2 )−g(a))+2

( +1)(+2),

and ∫ a+mb2

a((g(mb)−g(t))− (g(t)−g(a)))(g(t)−g(a))d(g(t))

=∫ mb

a+mb2

((g(t)−g(a))− (g(mb)−g(t)))(g(mb)−g(t))d(g(t))

= −(g( a+mb

2 )−g(a))+1

+1

(g(mb)−g(

a+mb2

))

+(g(mb)−g(a))+2

( +1)(+2)

−(g( a+mb

2 )−g(a))+2

( +1)(+2)−(g(mb)−g( a+mb

2 ))+2

+2.

Using the above evaluations of integrals in (8.43), we get the required inequality (8.39).�

Remark 8.7

(i) In Theorem 8.7, if we put m = 1, we get [132, Theorem ]

(ii) In Theorem 8.7, if we put g = I and p = 0, we get [3, Theorem 2.3].

(iii) In Theorem 8.7, if we put g = I, p = 0 and m = 1, we get [1, Theorem 2.3].

(iv) In Theorem 8.7, if we put g = I, we get [57, Theorem ].

(v) In Theorem 8.7, if we put g = I, m = 1, we get [33, Theorem 2.3].

(vi) In Theorem 8.7, for w = p = 0, g = I and h = 1 along with = m = 1, we get [42,Theorem 2.2].

(vii) In Theorem 8.7, if we put w = p = 0, g = I and h = 1 with m = 1, then we get [140,Theorem 3].


Theorem 8.8 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,g,h : [a,mb] → R, 0 < a < mb, Range (g), Range (h) ⊂ [a,mb] be the functions suchthat f be positive, ( f ◦g)′ ∈ L1[a,mb], g be a differentiable and strictly increasing and h becontinuous. Also let h(g(t)) = h(g(a)+g(mb)−g(t)) and |( f ◦g)′|q1 , q1 ≥ 1 is m-convex.Then for k < r +(), the following inequality for fractional integral operators (2.23)and (2.24) holds:∣∣∣∣

(f (g(a))+ f (g(mb))

2

)[(g

,r,k,c , , ,w,a+h◦g

)(mb; p)+

(g

,r,k,c , , ,w,mb−

h◦g)

(a; p)]

−[(

g ,r,k,c , , ,w,a+( f ◦g)(h◦g)

)(mb; p)+

(g

,r,k,c , , ,w,mb−

( f ◦g)(h◦g))

(a; p)]∣∣∣

≤ ‖ h ‖ M(g(mb)−g(a))+1

(+1)

((1−)

1p1 (1−)

1q1

)( | f ′(g(a))|q1 +m| f ′(g(b))|q1

2

) 1q1

,

(8.44)

where ‖ h ‖= supt∈[a,mb]

|h(t)|, 1p1

+ 1q1

= 1,

=(

g(mb)−g( a+mb2 )

g(mb)−g(a)

)+1

+(

g( a+mb2 )−g(a)

g(mb)−g(a)

)+1

and

= 1+2

[{(g( a+mb

2 )−g(a)g(mb)−g(a)

)+2}

+

{(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+2}]

− +1+2

[{(g( a+mb

2 )−g(a)g(mb)−g(a)

)+2}

+

{(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+2}]

−(

g( a+mb2 )−g(a)

g(mb)−g(a)

)+1(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+(

g( a+mb2 )−g(a)

g(mb)−g(a)

)(g(mb)−g( a+mb

2 )g(mb)−g(a)

)+1

.

Proof. Using Lemma 8.6, power mean inequality, (8.42) and m-convexity of |( f ◦ g)′|q1

respectively, we have∣∣∣∣(

f (g(a))+ f (g(mb))2

)[(g

,r,k,c , , ,w,a+h◦g

)(mb; p)+

(g

,r,k,c , , ,w,mb−

h◦g)

(a; p)]

−[(

g ,r,k,c , , ,w,a+( f ◦g)(h◦g)

)(mb; p)+

(g

,r,k,c , , ,w,mb−

( f ◦g)(h◦g))

(a; p)]∣∣∣

≤[∫ mb

a

∣∣∣∣∫ a+mb−t

t(g(mb)−g(s))−1E ,r,k,c

, , (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))∣∣∣∣d(g(t))

]1− 1q1

[∫ mb

a

∣∣∣∣∫ a+mb−t

t(g(mb)−g(s))−1E ,r,k,c

, , (w(g(mb)−g(s)) ; p)h(g(s))d(g(s))∣∣∣∣ | f ′(g(t))|q1

] 1q1

.

(8.45)

Since |( f ◦ g)′|q1 is m-convex, we have

| f ′(g(t))|q1 ≤ g(mb)−g(t)g(mb)−g(a)

| f ′(g(a))|q1 +mg(t)−g(a)

g(mb)−g(a)| f ′(g(b))|q1 . (8.46)

Using ‖ h ‖= supt∈[a,mb]

|h(t)|, and absolute convergence of extended Mittag-Leffler function,

inequality (8.45) becomes


∣∣∣∣(

f (g(a))+ f (g(mb))2

)[(g

,r,k,c ,, ,w,a+h ◦ g

)(mb; p)+

(g

,r,k,c ,, ,w,mb−h ◦ g

)(a; p)

]−[(

g ,r,k,c ,, ,w,a+( f ◦ g)(h ◦ g)

)(mb; p)+

(g

,r,k,c ,, ,w,mb−( f ◦ g)(h ◦ g)

)(a; p)

]∣∣∣≤ ‖ h ‖1− 1

q1 M1− 1

q1

[∫ a+mb2

a

(∫ a+mb−t

t(g(mb)−g(s))−1d(g(s))

)d(g(t))

+∫ b

a+mb2

(∫ t

a+mb−t(g(mb)−g(s))−1d(g(s))

)d(g(t))

]1− 1q1

× ‖ h ‖1q1 M

1q1

[∫ a+mb2

a

(∫ a+mb−t

t(g(mb)−g(s))−1d(g(s))

)

×(


| f ′(g(a))|q1 +mg(t)−g(a)

g(mb)−g(a)| f ′(g(b))|q1

)d(g(t))

+∫ b

a+mb2

(∫ t

a+mb−t(g(mb)−g(s))−1d(g(s))

)

×(


| f ′(g(a)))|q1 +mg(t)−g(a)

g(mb)−g(a)| f ′(g(b))|q1

)d(g(t))

] 1q1

.

After integrating and simplifying above inequality, we get (8.44). �

Remark 8.8

(i) In Theorem 8.8, if we put m = 1, we get [132, Theorem ].

(ii) In Theorem 8.8, if we put g = I and p = 0, we get [3, Theorem 2.6].

(iii) In Theorem 8.8, if we put g = I, p = 0 and m = 1 we get [1, Theorem 2.6].

(iv) In Theorem 8.8, if we put g = I, we get [57, Theorem ].

(v) In Theorem 8.8, if we put g = I,m = 1 we get [33, Theorem 2.5].


8.3 Bounds of Fractional Integral Operatorsfor (h−m)-convex Functions

In this section the bounds of extended generalized fractional integral operators for (h−m)-convex functions are given.

Theorem 8.9 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R, 0 ≤ a < mb, be a real valued function. If f is positive and (h−m)-convex,then for , ′ ≥ 1, the following inequality for extended generalized fractional integraloperators holds(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.47)

≤((x−a) f (a)Hw

a+,−1(x; p)+ (b− x) f (b)Hwb−, ′−1(x; p)

+mf( x

m

)((x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

))∫ 1

0h(z)dz.

Proof. Let x∈ [a,b]. Then first we observe the function f on the interval [a,x]; for t ∈ [a,x]and ≥ 1, we have the following inequality:

(x− t)−1E ,c,q,r , , (w(x− t) ; p) ≤ (x−a)−1E ,c,q,r

, , (w(x−a) ; p). (8.48)

As f is (h−m)-convex, for t ∈ [a,x], we have

f (t) ≤ h

(x− tx−a

)f (a)+mh

(t−ax−a

)f( x

m

). (8.49)

Multiplying (8.48) and (8.49), then integrating over [a,x], we get∫ x


, , (w(x− t) ; p) f (t)dt (8.50)

≤ f (a)(x−a)−1E ,c,q,r , , (w(x−a) ; p)

∫ x

ah

(x− tx−a

)dt

+mf( x

m

)(x−a)−1E ,c,q,r

, , (w(x−a) ; p)∫ x

ah

(t−ax−a

)dt.

By using (2.2), we have

(w, ,c,q,ra+, , , f

)(x; p) ≤ (x−a)Hw

a+,−1(x; p)(

f (a)+mf( x

m

))∫ 1

0h(z)dz. (8.51)

Now on the other hand we observe the function f on the interval [x,b]; for t ∈ [x,b] and ′ ≥ 1, we get the following inequality:

(t − x)′−1E ,c,q,r

, ′, (w(t− x) ; p) ≤ (b− x)′−1E ,c,q,r

, ′, (w(b− x) ; p). (8.52)

8.3 BOUNDS FOR (h−m)-CONVEX FUNCTIONS 155

Again from (h−m)-convexity of f for t ∈ [x,b], we have

f (t) ≤ h

(t− xb− x

)f (b)+mh

(b− tb− x

)f( x

m

). (8.53)

Similarly multiplying (8.52) and (8.53), then integrating over [x,b], we get

(w, ,c,q,rb−, , ′, f

)(x; p) ≤ (b− x)Hw

b−, ′−1(x; p)(

f (b)+mf( x

m

))∫ 1

0h(z)dz. (8.54)

Adding (8.51) and (8.54), inequality (8.47) is obtained. �

If m = 1 and h(z) = z in (8.47), then the following result holds for convex functions:

Corollary 8.3 Let f : [a,b] → R, a < b, be a real valued function. If f is positive andconvex, then for , ′ ≥ 1, we have(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.55)

≤(x−a) f (a)Hw

a+,−1(x; p)+ (b− x) f (b)Hwb−, ′−1(x; p)

2

+ f (x)[ (x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

2

].

Remark 8.9 If w = p = 0 in (8.55), then [47, Theorem 1] is obtained.

Lemma 8.7 [45] Let f : [0,) → R be a (h−m)-convex function. If 0 ≤ a < mb andf (x) = f

(a+b−x

m

), then the following inequality holds:

f

(a+b

2

)≤ (m+1)h

(12

)f (x), x ∈ [a,b]. (8.56)

Theorem 8.10 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R, 0 ≤ a < b, be a real valued function. If f is positive, (h−m)-convex andf (x) = f

(a+b−x

m

), then for , ′ > 0,the following inequality for extended generalized

fractional integral operators holds

f(

a+b2

)(m+1)h

(12

) [Hwb−, ′+1(a; p)+Hw

a+,+1(b; p)]

(8.57)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤ (b−a)2[Hw

b−, ′−1(a; p)+Hwa+,−1(b; p)

](f (a)+mf

(bm

))∫ 1

0h(z)dz.

Proof. For x ∈ [a,b], we have

(x−a)′E ,c,q,r , ′, (w(x−a) ; p) ≤ (b−a)

′E ,c,q,r , ′, (w(b−a) ; p), ′ > 0. (8.58)


As f is (h−m)-convex, for x ∈ [a,b] we have

f (x) ≤ mh

(x−ab−a

)f

(bm

)+h

(b− xb−a

)f (a). (8.59)

Multiplying (8.58) and (8.59), then integrating over [a,b], we get

∫ b

a(x−a)

′E ,c,q,r , ′, (w(x−a) ; p) f (x)dx

≤ m(b−a)′E ,c,q,r , ′, (w(b−a) ; p) f

(bm

)∫ b

ah

(x−ab−a

)dx

+(b−a)′E ,c,q,r , ′, (w(b−a) ; p) f (a)

∫ b

ah

(b− xb−a

)dx.

From which we have

(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)2Hw

b−, ′−1(a; p)(

f (a)+mf

(bm

))∫ 1

0h(z)dz. (8.60)

On the other hand for x ∈ [a,b], we have

(b− x)E ,c,q,r , , (w(b− x) ; p) ≤ (b−a)E ,c,q,r

, , (w(b−a) ; p) > 0. (8.61)

Similarly multiplying (8.59) and (8.61), then integrating over [a,b], we get

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)2Hw

a+,−1(b; p)(

f (a)+mf

(bm

))∫ 1

0h(z)dz. (8.62)

Adding (8.60) and (8.62), we get(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤ (b−a)2(Hw

b−, ′−1(a; p)+Hwa+,−1(b; p)

)(f (a)+mf

(bm

))∫ 1

0h(z)dz.

Multiplying (8.56) with (x−a)′E ,c,q,r , ′, (w(x−a) ; p), then integrating over [a,b], we get

f

(a+b

2

)∫ b

a(x−a)

′E ,c,q,r , ′, (w(x−a) ; p)dx

≤ (m+1)h(

12

)∫ b

a(x−a)

′E ,c,q,r , ′, (w(x−a) ; p) f (x)dx,

by using (2.2), we get

f(

a+b2

)(m+1)h

(12

)Hwb−, ′+1(a; p) ≤

(w, ,c,q,rb−, , ′+1, f

)(a; p). (8.63)


Similarly multiplying (8.56) with (b−x)E ,c,q,r , , (w(b−x) ; p), then integrating over [a,b]

and using (2.2), we get

f(

a+b2

)(m+1)h

(12

)Hwa+,+1(b; p) ≤

(w, ,c,q,ra+, ,+1, f

)(b; p). (8.64)

Adding (8.63) and (8.64), we get

f(

a+b2

)(m+1)h

(12

) [Hwb−, ′+1(a; p)+Hw

a+,+1(b; p)]

(8.65)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p).

From inequalities (8.63) and (8.65), inequality (8.57) is obtained. �

If m = 1 and h(z) = z in (8.57), then the following result holds for convex function:

Corollary 8.4 Let f : [a,b]−→R, a < b, be a real valued function. If f is positive, convexand symmetric about a+b

2 , then for , ′ > 0, we have

f

(a+b

2

)[Hw

b−, ′+1(a; p)+Hwa+,+1(b; p)

](8.66)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤ (b−a)2[Hw

b−, ′−1(a; p)+Hwa+,−1(b; p)

][ f (a)+ f (b)2

].

Remark 8.10

(i) If w = p = 0 in (8.66), then [47, Theorem 3] is obtained.

(ii) If , ′ → 0 and w = p = 0, then from above inequality, we get the Hadamardinequality.

Theorem 8.11 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r+ . Let f :[a,b]→ R, 0 ≤ a < mb, be a real valued function. If f is differentiable and | f ′| is (h−m)-convex, then for , ′ ≥ 1, the following inequality for extended generalized fractionalintegral operators holds∣∣∣(w, ,c,q,r

a+, ,−1, f)

(x; p)+(w, ,c,q,rb−, , ′−1, f

)(x; p) (8.67)

−(

f (a)Hwa+,−1(x; p)+ f (b)Hw

b−, ′−1(x; p))∣∣∣

≤((x−a)| f ′(a)|Hw

a+,−1(x; p)+ (b− x)| f ′(b)|Hwb−, ′−1(x; p)

+m∣∣∣ f ′( x

m

)∣∣∣((x−a)Hwa+,−1(x; p)+ (b− x)Hw

b−, ′−1(x; p)))∫ 1

0h(z)dz.


Proof. Let x ∈ [a,b] and t ∈ [a,x]. Then using (h−m)-convexity of | f ′|, we have

| f ′(t)| ≤ h

(x− tx−a

)| f ′(a)|+mh

(t−ax−a

)∣∣∣ f ′( xm

)∣∣∣ . (8.68)

From (8.68) follows

f ′(t) ≤ h

(x− tx−a

)| f ′(a)|+mh

(t−ax−a

)∣∣∣ f ′( xm

)∣∣∣ . (8.69)

Multiplying (8.48) and (8.69), then integrating over [a,x], we get

∫ x


, , (w(x− t) ; p) f ′(t)dt (8.70)

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)

×[| f ′(a)|

∫ x

ah

(x− tx−a

)dt +m

∣∣∣ f ′( xm

)∣∣∣∫ x

ah

(t−ax−a

)dt

].

The left hand side is calculated as:∫ x


, , (w(x− t) ; p) f ′(t)dt, (8.71)

put x− t = z that is t = x− z, also using the derivative property of Mittag-Leffler function,we have∫ x−a

0z−1E ,c,q,r

, , (wz ; p) f ′(x− z)dz

= (x−a)−1E ,c,q,r , , (w(x−a) ; p) f (a)−

∫ x−a

0z−2E ,c,q,r

,−1,(wz ; p) f (x− z)dz,

now put x− z = t in second term of the right hand side of the above equation, we get

∫ x


, , (w(x− t) ; p) f ′(t)dt (8.72)

=∫ x−a

0z−1E ,c,q,r

, , (wz ; p) f ′(x− z)dz

= (x−a)−1E ,c,q,r , , (w(x−a) ; p) f (a)−

(w, ,c,q,ra+, ,+1, f

)(x; p)

Hwa+,−1(x; p) f (a)−

(w, ,c,q,ra+, ,+1, f

)(x; p).

Therefore, (8.70) takes the form as follows:

f (a)Hwa+,−1(x; p)−

(w, ,c,q,ra+, ,−1, f

)(x; p) (8.73)

≤ (x−a)Hwa+,−1(x; p)

(| f ′(a)|+m

∣∣∣ f ′( xm

)∣∣∣)∫ 1

0h(z)dz.


Also, from (8.68) we obtain

f ′(t) ≥−(

h

(x− tx−a

)| f ′(a)|+mh

(t −ax−a

)∣∣∣ f ′( xm

)∣∣∣) . (8.74)

Following the same procedure as did for (8.69), we also have(w, ,c,q,ra+, ,−1, f

)(x; p)− f (a)Hw

a+,−1(x; p) (8.75)

≤ (x−a)Hwa+,−1(x; p)

(| f ′(a)|+m

∣∣∣ f ′( xm

)∣∣∣)∫ 1

0h(z)dz.

From (8.73) and (8.75), we get∣∣∣(w, ,c,q,ra+, ,−1, f

)(x; p)− f (a)Hw

a+,−1(x; p)∣∣∣ (8.76)

≤ (x−a)Hwa+,−1(x; p)

(| f ′(a)|+m

∣∣∣ f ′( xm

)∣∣∣)∫ 1

0h(z)dz.

Now let t ∈ [x,b]. Then using (h−m)-convexity of | f ′|, we have

| f ′(t)| ≤ h

(t − xb− x

)| f ′(b)|+mh

(b− tb− x

)∣∣∣ f ′( xm

)∣∣∣ . (8.77)

On the same lines as we have done for (8.48), (8.69) and (8.74), from (8.52) and (8.77) wehave the following inequality:∣∣∣(w, ,c,q,r

b−, , ′−1, f)

(x; p)− f (b)Hwb−, ′−1(x; p)

∣∣∣ (8.78)

≤ (b− x)Hwb−, ′−1(x; p)

(| f ′(b)|+m

∣∣∣ f ′( xm

)∣∣∣)∫ 1

0h(z)dz.

From (8.76) and (8.78) via triangular inequality, inequality (8.67) is obtained. �

If m = 1 and h(z) = z in (8.67), then the following result holds for convex function:

Corollary 8.5 Let f : [a,b] −→ R, a < b, be a real valued function. If f is differentiableand | f ′| is convex, then for , ′ ≥ 1, we have∣∣∣(w, ,c,q,r

a+, ,−1, f)

(x; p)+(w, ,c,q,rb−, , ′−1, f

)(x; p) (8.79)

−(

f (a)Hwa+,−1(x; p)+ f (b)Hw

b−, ′−1(x; p))∣∣∣

≤(x−a)| f ′(a)|Hw

a+,−1(x; p)+ (b− x)| f ′(b)|Hwb−, ′−1(x; p)

2

+| f ′(x)|(

(x−a)Hwa+,−1(x; p)+ (b− x)Hw

b−, ′−1(x; p)

2

).

Remark 8.11

(i) If w = p = 0 and replace by +1 in (8.79), then [47, Theorem 2] is obtained.

(ii) If w = p = 0, = = 1 and f ′ passes through x = a+b2 , then from (8.79) [42,

Theorem 2.2] is obtained.


8.4 Inequalities for the Extended GeneralizedMittag-Leffler Function

In this section, results of previous section are applied for the function f (x) = x2. Function fis convex and | f ′(x)| = 2|x| which is also convex. By virtue of this function we succeededto establish recurrence relations among Mittag-Leffler functions useful in the solutionsof fractional boundary value problems and fractional differential equations. Ullah et al.computed generalized fractional integral operators for the function f (x) = x2, in [148], asfollows: (

w, ,c,q,ra+, , , f

)(x; p) = (x−a)

[a2E ,c,q,r

,+1,(w(x−a) ; p) (8.80)

+2a(x−a)E ,c,q,r ,+2,(w(x−a) ; p)+2(x−a)2E ,c,q,r

,+3,(w(x−a) ; p)].

(w, ,c,q,rb−, , , f

)(x; p) = (b− x)

[b2E ,c,q,r

,+1,(w(b− x) ; p) (8.81)

−2b(b− x)E ,c,q,r ,+2,(w(b− x) ; p)+2(b− x)2E ,c,q,r

,+3,(w(b− x) ; p)].

Theorem 8.12 Mittag-Leffler functions satisfy the following recurrence relation:

(a2 +b2)(b−a)2 E ,c,q,r

,+1,(w′(b−a) ; p)+E ,c,q,r

,+3,(w′(b−a) ; p) (8.82)

≤ ((2m+1)(a2 +b2)+2ab)2m(b−a)2 E ,c,q,r

, , (w′(b−a) ; p)∫ 1

0h(z)dz+E ,c,q,r

,+2,(w′(b−a) ; p)

where w′ = w2 .

Proof. By using (8.80), (8.81) and f (x) = x2 in (8.47) of Theorem 8.9, we have

(x−a)[a2E ,c,q,r

,+1,(w(x−a) ; p)+2a(x−a)E ,c,q,r ,+2,(w(x−a) ; p) (8.83)

+2(x−a)2E ,c,q,r ,+3,(w(x−a) ; p)

]+(b− x)

′ [b2E ,c,q,r

, ′+1,(w(b− x) ; p)

−2b(b− x)E ,c,q,r , ′+2,(w(b− x) ; p)+2(b− x)2E ,c,q,r

, ′+3,(w(b− x) ; p)]

≤((

a2 +x2

m

)(x−a)E ,c,q,r

, , (w(x−a) ; p)

+(

b2 +x2

m

)(b− x)

′E ,c,q,r , ′, (w(b− x) ; p)

)∫ 1

0h(z)dz.

Now by putting x = a+b2 and = ′ in (8.83), then after simplification, inequality

(8.82) is obtained. �

8.4 INEQUALITIES FOR THE GENERALIZED MITTAG-LEFFLER FUNCTION 161

Corollary 8.6 If m = 1 and h(z) = z in (8.82), then we have

(a2 +b2)(b−a)2 E ,c,q,r

,+1,(w′(b−a) ; p)+E ,c,q,r

,+3,(w′(b−a) ; p) (8.84)

≤ (3a2 +3b2 +2ab)4(b−a)2 E ,c,q,r

, , (w′(b−a) ; p)+E ,c,q,r ,+2,(w

′(b−a) ; p).

Theorem 8.13 Mittag-Leffler functions satisfy the following recurrence relation:∣∣∣E ,c,q,r ,+2,(w

′(b−a); p)−E ,c,q,r ,+1,(w

′(b−a) ; p)∣∣∣ (8.85)

≤ 1m(b−a)

(ma+mb+(a+b))E ,c,q,r , , (w′(b−a) ; p)

∫ 1

0h(z)dz

where w′ = w2 .

Proof. By using (8.80), (8.81) and | f ′(x)| = 2|x| in (8.67) of Theorem 8.11, we have∣∣∣(x−a)[a2(x−a)−1E ,c,q,r

, , (w(x−a) ; p)+2aE ,c,q,r ,+1,(w(x−a) ; p) (8.86)

+2(x−a)E ,c,q,r ,+2,(w(x−a) ; p)

]+(b− x)

′[b2(b− x)−1E ,c,q,r

, ′, (w(b− x) ; p)

−2bE ,c,q,r , ′+1,(w(b− x) ; p)+2(b− x)E ,c,q,r

, ′+2,(w(b− x) ; p)]

−(a2(x−a)−1E ,c,q,r

, , (w(x−a) ; p)+b2(b− x)′−1E ,c,q,r

, ′, (w(b− x) ; p))∣∣∣

≤(

2

(|a|+ |x|

m

)(x−a)E ,c,q,r

, , (w(x−a) ; p)

+2

(|b|+ |x|

m

)(b− x)

′E ,c,q,r , ′, (w(b− x) ; p)

)∫ 1

0h(z)dz.

Now by putting x = a+b2 and = ′ in (8.86), then after simplification, inequality (8.85)

is obtained. �

Corollary 8.7 If m = 1 and h(z) = z in (8.85), then we have∣∣∣E ,c,q,r ,+2,(w

′(b−a) ; p)−E ,c,q,r ,+1,(w

′(b−a) ; p)∣∣∣ (8.87)

≤ 1b−a

(a+b)E ,c,q,r , , (w′(b−a) ; p).


E ,c,q,r ,+1,(w(b−a) ; p)− (1+

1m

)E ,c,q,r , , (w(b−a) ; p)

∫ 1

0h(z)dz (8.88)

≤ 2(b−a)2

(a2 +b2)

(E ,c,q,r ,+2,(w(b−a) ; p)−2E ,c,q,r

,+3,(w(b−a) ; p))

.


Proof. In (8.83), putting x = a and x = b, then adding for = ′, inequality (8.88) isobtained. �


E ,c,q,r ,+1,(w(b−a) ; p)−E ,c,q,r

, , (w(b−a) ; p) (8.89)

≤ 2(b−a)2

(a2 +b2)

(E ,c,q,r ,+2,(w(b−a) ; p)−2E ,c,q,r

,+3,(w(b−a) ; p))

.

Theorem 8.15 Mittag-Leffler functions satisfy the following recurrence relation:∣∣∣∣E ,c,q,r ,+2,(w(b−a) ; p)− 1

2E ,c,q,r ,+1,(w(b−a) ; p)

∣∣∣∣ (8.90)

≤ (1+ 1m)

2(b−a)(a+b)E ,c,q,r

, , (w(b−a) ; p)∫ 1

0h(z)dz.

Proof. In (8.86), putting x = a and x = b, then adding for = ′, inequality (8.90) isobtained. �

Corollary 8.9 If m = 1 and h(z) = z in (8.90), then we have∣∣∣∣E ,c,q,r ,+2,(w(b−a) ; p)− 1

2E ,c,q,r ,+1,(w(b−a) ; p)

∣∣∣∣ (8.91)

≤ 12(b−a)

(a+b)E ,c,q,r , , (w(b−a) ; p).


E ,c,q,r ,+1,(w; p)− (1+

1m

)E ,c,q,r , , (w; p)

∫ 1

0h(z)dz (8.92)

≤ 2(E ,c,q,r ,+2,(w; p)−2E ,c,q,r

,+3,(w; p))

.

Proof. In (8.88), putting a = 0 and b = 1, then inequality (8.92) is obtained. �


E ,c,q,r ,+1,(w; p)−E ,c,q,r

, , (w; p) ≤ 2(E ,c,q,r ,+2,(w; p)−2E ,c,q,r

,+3,(w; p))

. (8.93)

Theorem 8.17 Mittag-Leffler functions satisfy the following recurrence relation:∣∣∣2E ,c,q,r ,+2,(w; p)−E ,c,q,r

,+1,(w; p)∣∣∣≤ (1+

1m

)E ,c,q,r , , (w; p)

∫ 1

0h(z)dz. (8.94)

Proof. In (8.90), putting a = 0 and b = 1, then inequality (8.94) is obtained. �

Corollary 8.11 If m = 1 and h(z) = z in (8.94), then we have∣∣∣2E ,c,q,r ,+2,(w; p)−E ,c,q,r

,+1,(w; p)∣∣∣≤ E ,c,q,r

, , (w; p). (8.95)

By applying Theorem 8.10 similar relations can be established which we leave for thereader.

8.5 ERROR BOUNDS FOR QUASI-CONVEX FUNCTIONS 163

8.5 Error Bounds of Fractional Integral Operatorsfor Quasi-convex Functions

The very first result provides the upper bound of the sum of the left sided and right sidedfractional integrals via quasi-convex functions.

Theorem 8.18 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b]→ R be a function such that f ∈ L1[a,b] with a < b. If f is quasi-convex on [a,b],then the following inequality holds(w′, ,c,q,ra+, , , f

)(b; p)+

(w′, ,c,q,rb−, , ′, f

)(a; p) ≤ max{ f (a), f (b)}

(Hw′

a+, (b; p)+Hw′b−, ′(a; p)

)(8.96)


Proof. By using quasi-convexity in the form f ((1− t)a + tb) ≤ max{ f (a), f (b)} thefollowing inequality can be obtained∫ 1

0t−1E ,c,q,r

, , (wt ; p) f (ta+(1− t)b)dt ≤ max{ f (a), f (b)}∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt.

(8.97)

Making substitution x = ta+(1− t)b that is t = b−xb−a in the inequality (8.97). Then it takes

the form ∫ b

a

(b− xb−a

)−1

E ,c,q,r , ,

(w

(b− xb−a

); p

)f (x)

dxb−a

(8.98)

≤ max{ f (a), f (b)}∫ b

a

(b− xb−a

)−1

E ,c,q,r , ,

(w

(b− xb−a

); p

)dx

b−a.

From above inequality follows(w′, ,c,q,ra+, , , f

)(b; p) ≤ Hw′

a+, (b; p)max{ f (a), f (b)}. (8.99)

Also by using quasi-convexity in the form f ((1− t)a+ tb)≤ max{ f (a), f (b)} the follow-ing inequality can be obtained∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ((1− t)a+ tb)dt ≤ max{ f (a), f (b)}∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt.

(8.100)

Now making substitution of y = (1− t)a+ tb that is t = y−ab−a in the inequality (8.100). Then

it takes the form∫ b

a

(y−ab−a

)−1

E ,c,q,r , ,

(w

(y−ab−a

); p

)f (y)

dyb−a

(8.101)

≤ max{ f (a), f (b)}∫ b

a

(y−ab−a

)−1

E ,c,q,r , ,

(w

(y−ab−a

); p

)dy

b−a.


From above inequality we obtain(w′, ,c,q,rb−, , ′, f

)(a; p) ≤ Hw′

b−, ′(a; p)max{ f (a), f (b)}. (8.102)

Adding (8.99) and (8.102), we get (8.96) which is required. �

Corollary 8.12 Setting = ′ in (8.96), then we get the following inequality(w′, ,c,q,ra+, , , f

)(b; p)+

(w′ , ,c,q,rb−, , ′, f

)(a; p)≤ max{ f (a), f (b)}

(Hw′

a+, (b; p)+Hw′b−, (a; p)

).

(8.103)

Corollary 8.13 [118] Setting w = p = 0 in (8.103), then we get the following inequalityfor Riemann-Liouville fractional integrals

Ja+ f (b)+ Jb− f (a) ≤ 2(b−a)

( +1)max{ f (a), f (b)}. (8.104)

Remark 8.12 If we take = 1 in (8.104), then we get the following inequality for quasi-convex function which is related to the Hadamard inequality given by Dragomir and Pearcein [44]

1b−a

∫ b

af (t)dt ≤ max{ f (a), f (b)}. (8.105)

8.5.1 Recurrence Inequalities for Mittag-Leffler Functions

Let us consider the function f (x) = x2. The function f is convex on [a,b] and | f ′(x)|= 2|x|which is again convex function on [a,b]. Since f and | f ′| are convex and finite on [a,b],therefore are quasi-convex. Results of previous section are applied for this function andinequalities among the generalized extended Mittag-Leffler function are established.

Theorem 8.19 The Mittag-Leffler function satisfies the following recurrence inequality

2E ,c,q,r ,+3,(w

′(b−a) ; p)−E ,c,q,r ,+2,(w

′(b−a) ; p) (8.106)

≤ E ,c,q,r ,+1,(w

′(b−a) ; p)[max{a2,b2}− (a2 +b2)

2(b−a)2

]


Proof. For the function f (t) = t2 the generalized fractional integral operator is evaluatedas follows(

w, ,c,q,ra+, , , f

)(x; p) (8.107)

=∫ x


, , (w(x− t) ; p)t2dt

=∫ x

a(x− t)−1

n=0

p( +nq,r− ) ( ,r− )

(r)nq

(n+)wn(x− t)n

()nct2dt


=

n=0

p( +nq,r− ) ( ,r− )

(r)nq

(n+)wn

()nc

∫ x

a(x− t)n+−1t2dt

= (x−a)[a2E ,c,q,r

,+1,(w(x−a) ; p)+2a(x−a)E ,c,q,r ,+2,(w(x−a) ; p)

+2(x−a)2E ,c,q,r ,+3,(w(x−a) ; p)

],

Similarly(w, ,c,q,rb−, , ,, f

)(x; p) = (b− x)

[b2E ,c,q,r

,+1,(w(b− x) ; p)−2b(b− x) (8.108)

E ,c,q,r ,+2,(w(b− x) ; p)+2(b− x)2E ,c,q,r

,+3,(w(b− x) ; p)].

Using (8.96) of Theorem 8.18 for the function t2 takes the form

(b−a)[a2E ,c,q,r

,+1,(w′(b−a) ; p)+2a(b−a)E ,c,q,r

,+2,(w′(b−a) ; p) (8.109)

+2(b−a)2E ,c,q,r ,+3,(w

′(b−a) ; p)]+(b−a)

′ [b2E ,c,q,r

, ′+1,(w′(b−a) ; p)

−2b(b−a)E ,c,q,r , ′+2,(w

′(b−a) ; p)+2(b−a)2E ,c,q,r , ′+3,(w

′(b−a) ; p)]

≤ 2max{a2,b2}((b−a)E ,c,q,r

,+1,(w′(b−a) ; p)

+(b−a)′E ,c,q,r , ′+1,(w

′(b−a) ; p))

.

Now taking = ′ in (8.109), then after simplification we get (8.106). �

Theorem 8.20 The Mittag-Leffler function satisfies the following recurrence inequality∣∣∣∣∣a2 +b2

2− 1

2E ,c,q,r , , (w; p)

[(a2 +b2)E ,c,q,r

, , (w′(b−a) ; p) (8.110)

+2(b−a)2(2E ,c,q,r ,+2,(w

′(b−a) ; p)−E ,c,q,r ,+1,(w

′(b−a) ; p))]∣∣∣

≤ 2(b−a)M

E ,c,q,r , , (w; p)

max{|a|, |b|}


Proof. By using (8.107), (8.108) and f (t) = t2, | f ′(t)| = 2|t| in (8.119) of Theorem 8.23,we have ∣∣∣∣∣a

2 +b2

2− 1

2(b−a)−1E ,c,q,r , , (w; p)

(8.111)

×[(b−a)

((b−a)−1(a2 +b2)E ,c,q,r

, , (w′(b−a) ; p)


−2(b−a)E ,c,q,r ,+1,(w

′(b−a) ; p)+4(b−a)E ,c,q,r ,+2,(w

′(b−a) ; p))]∣∣∣

≤ (b−a)M

E ,c,q,r , , (w; p)

max{2|a|,2|b|}.

After simplification we get (8.110). �


2− 1

2E ,c,q,r , , (w; p)

[(a2 +b2)E ,c,q,r

, , (w′(b−a) ; p) (8.112)

+2(b−a)2(2E ,c,q,r ,+2,(w

′(b−a) ; p)−E ,c,q,r ,+1,(w

′(b−a) ; p))]∣∣∣

≤ 2(b−a)M

E ,c,q,r , , (w; p)

1

(( −1)p+1)1p

(max{|a|q, |b|q}) 1q


Proof.By using (8.107), (8.108) and f (t) = t2, | f ′(t)| = 2|t| in (8.123) of Theorem 8.24, we

have ∣∣∣∣∣a2 +b2

2− 1

2(b−a)−1E ,c,q,r , , (w; p)

(8.113)

×[(b−a)

((b−a)−1(a2 +b2)E ,c,q,r

, , (w′(b−a) ; p)

−2(b−a)E ,c,q,r ,+1,(w

′(b−a) ; p)+4(b−a)E ,c,q,r ,+2,(w

′(b−a) ; p))]∣∣∣

≤ (b−a)M

E ,c,q,r , , (w; p)

1

(( −1)p+1)1p

(max{(2|a|)q,(2|b|)q}) 1q .



2− 1

2E ,c,q,r , , (w; p)

[(a2 +b2)E ,c,q,r

, , (w′(b−a) ; p) (8.114)

+2(b−a)2(2E ,c,q,r ,+2,(w

′(b−a) ; p)−E ,c,q,r ,+1,(w

′(b−a) ; p))]∣∣∣

≤ 2(b−a)M

E ,c,q,r , , (w; p)

(max{|a|q, |b|q}) 1q



Proof. By using (8.107), (8.108) and f (t) = t2, | f ′(t)| = 2|t| in (7.29) of Theorem 8.25,we have ∣∣∣∣∣a

2 +b2

2− 1

2(b−a)−1E ,c,q,r , , (w; p)

(8.115)

×[(b−a)

((b−a)−1(a2 +b2)E ,c,q,r

, , (w′(b−a) ; p)

−2(b−a)E ,c,q,r ,+1,(w

′(b−a) ; p)+4(b−a)E ,c,q,r ,+2,(w

′(b−a) ; p))]∣∣∣

≤ 2(b−a)M

E ,c,q,r , , (w; p)

(max{|a|q, |b|q}) 1q .


8.5.2 Error Bounds of Hadamard andFejer-Hadamard Inequalities

The following identity is very important to give the Hadamard type inequalities.

Lemma 8.8 Let f : [a,b] → R be a function such that f ′ ∈ L1[a,b] with a < b, then forgeneralized fractional integral operators the following identity holds

f (a)+ f (b)2

−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′ , ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

(8.116)

=b−a

2E ,c,q,r , , (w; p)

∫ 1

0

((1− t)−1E ,c,q,r

, , (w(1− t) ; p)− t−1E ,c,q,r , , (wt ; p)

)× f ′(ta+(1− t)b)dt


Proof. We have

b−a

2E ,c,q,r , , (w; p)

∫ 1

0

((1− t)−1E ,c,q,r

, , (w(1− t) ; p)− t−1E ,c,q,r , , (wt ; p)

)(8.117)

× f ′(ta+(1− t)b)dt

=b−a

2E ,c,q,r , , (w; p)

∫ 1

0(1− t)−1E ,c,q,r

, , (w(1− t) ; p) f ′(ta+(1− t)b)dt

− b−a

2E ,c,q,r , , (w; p)

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ′(ta+(1− t)b)dt.

We first consider the first term of right hand side of (8.117): putting z = 1− t that ist = 1− z and using the derivative property of Mittag-Leffler function, it takes form


b−a

2E ,c,q,r , , (w; p)

∫ 1

0z−1E ,c,q,r

, , (wz ; p) f ′((1− z)a+ zb)dz =b−a

2E ,c,q,r , , (w; p)

×[

E ,c,q,r , , (w; p) f (b)

b−a− 1

b−a

∫ 1

0z−2E, ,k,c

,−1,(wz ; p) f ((1− z)a+ zb)dz

].

Making substitution x = (1− z)a+ zb in above we get

b−a

2E ,c,q,r , , (w; p)

∫ 1

0z−1E ,c,q,r

, , (wz ; p) f ′((1− z)a+ zb)dz (8.118)

=b−a

2E ,c,q,r , , (w; p)

[E ,c,q,r , , (w; p) f (b)

b−a− 1

(b−a)(w′ , ,c,q,rb−, ,−1, f

)(a; p)

].

Similarly consider the second term of right hand side of (8.117), we get

− b−a

2E ,c,q,r , , (w; p)

∫ 1

0t−1E ,c,q,r

, , (wt ; p) f ′(ta+(1− t)b)dt

= − b−a

2E ,c,q,r , , (w; p)

[−E ,c,q,r

, , (w; p) f (a)b−a

+1

(b−a)(w′, ,c,q,ra+, ,−1, f

)(b; p)

],

here we use substitution x = ta+(1− t)b.Now by using final form of both terms in (8.117), identity (8.116) is established. �

In the following we give Hadamard type inequality by using the above lemma.

Theorem 8.23 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b]→ R be a differentiable function such that f ′ ∈ L1[a,b] with a < b. If | f ′| is quasi-convex on [a,b], then for generalized fractional integral operators the following inequalityholds ∣∣∣∣∣∣

f (a)+ f (b)2

−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′ , ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.119)

≤ (b−a)M

E ,c,q,r , , (w; p)

max{| f ′(a)|, | f ′(b)|}

for q < r+(), where w′ = w(b−a) .

Proof. Using Lemma 8.8 and properties of modulus, we have∣∣∣∣∣∣f (a)+ f (b)

2−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′ , ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.120)

≤ b−a

2E ,c,q,r , , (w; p)

∫ 1

0

∣∣∣(1− t)−1E ,c,q,r , , (w(1− t) ; p)− t−1E ,c,q,r

, , (wt ; p)∣∣∣

×| f ′(ta+(1− t)b|dt


≤ b−a

2E ,c,q,r , , (w; p)

(∫ 1

0

∣∣∣(1− t)−1E ,c,q,r , , (w(1− t) ; p)

∣∣∣+∫ 1

0

∣∣∣t−1E ,c,q,r , , (wt ; p)

∣∣∣) | f ′(ta+(1− t)b|dt.

Since | f ′| is quasi-convex, also using absolute convergence of Mittag-Leffler function, wehave ∣∣∣∣∣∣

f (a)+ f (b)2

−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′, ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.121)

≤ (b−a)M

2E ,c,q,r , , (w; p)

(∫ 1

0|(1− t)−1|dt +

∫ 1

0|t−1|dt

)max{| f ′(a)|, | f ′(b)|}.

After simple calculation we get (8.119) which is required. �

Corollary 8.14 Setting w = p = 0 in (8.119), then we get the following inequality forRiemann-Liouville fractional integrals∣∣∣∣ f (a)+ f (b)

2− ()

2(b−a)−1

[J−1a+ f (b)+ J−1

b− f (a)]∣∣∣∣≤ b−a

max{| f ′(a)|, | f ′(b)|}.

(8.122)

Corollary 8.15 If we take = 2 in (8.122), then we get the following inequality forquasi-convex function∣∣∣∣ f (a)+ f (b)

2− 1

b−a

∫ b

af (t)dt

∣∣∣∣≤ b−a2

max{| f ′(a)|, | f ′(b)|}.

In the following we give the Hadamard type inequality by using Lemma 8.8, Holder’sinequality and quasi-convexity of | f ′|q.Theorem 8.24 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R be a differentiable function such that f ′ ∈ L1[a,b] with a < b. If | f ′|q, q > 1is quasi-convex on [a,b], then for generalized fractional integral operators the followinginequality holds

∣∣∣∣∣∣f (a)+ f (b)

2−(w′ , ,c,q,ra+, ,−1, f

)(b; p)+

(w′, ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.123)

≤ (b−a)M

E ,c,q,r , , (w; p)

1

(( −1)p+1)1p

(max{| f ′(a)|q, | f ′(b)|q}) 1

q

for q < r+() and 1p + 1

q = 1, where w′ = w(b−a) .


Proof. From Lemma 8.8, properties of modulus and Holder’s inequality, we have∣∣∣∣∣∣f (a)+ f (b)

2−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′ , ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.124)

≤ b−a

2E ,c,q,r , , (w; p)

∫ 1

0

∣∣∣(1− t)−1E ,c,q,r , , (w(1− t) ; p)− t−1E ,c,q,r

, , (wt ; p)∣∣∣

×| f ′(ta+(1− t)b|dt

≤ b−a

2E ,c,q,r , , (w; p)

(∫ 1

0

∣∣∣(1− t)−1E ,c,q,r , , (w(1− t) ; p)

∣∣∣+∫ 1

0

∣∣∣t−1E ,c,q,r , , (wt ; p)

∣∣∣) | f ′(ta+(1− t)b|dt

≤ b−a

2E ,c,q,r , , (w; p)

((∫ 1

0

∣∣∣(1− t)−1E ,c,q,r , , (w(1− t) ; p)

∣∣∣p dt

) 1p

+(∫ 1

0

∣∣∣t−1E ,c,q,r , , (wt ; p)

∣∣∣p dt

) 1p)(∫ 1

0| f ′(ta+(1− t)b|qdt

) 1q

.

Since | f ′|q is quasi-convex, also using absolute convergence of Mittag-Leffler function, wehave ∣∣∣∣∣∣

f (a)+ f (b)2

−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′, ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.125)

≤ (b−a)M

2E ,c,q,r , , (w; p)

((∫ 1

0

∣∣(1− t)−1∣∣p dt

) 1p

+(∫ 1

0

∣∣t−1∣∣p dt

) 1p)

× (max{| f ′(a)|q, | f ′(b)|q) 1q .



2− ()

2(b−a)−1

[J−1a+ f (b)+ J−1

b− f (a)]∣∣∣∣ (8.126)

≤ b−a

(( −1)p+1)1p

(max{| f ′(a)|q, | f ′(b)|q}) 1

q .


2− 1

b−a

∫ b

af (t)dt

∣∣∣∣≤ b−a

(p+1)1p

(max{| f ′(a)|q, | f ′(b)|q}) 1

q .


Theorem 8.25 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] → R be a differentiable function such that f ′ ∈ L1[a,b] with a < b. If | f ′|q, q ≥ 1is quasi-convex on [a,b], then for generalized fractional integral operators the followinginequality holds∣∣∣∣∣∣

f (a)+ f (b)2

−(w′ , ,c,q,ra+, ,−1, f

)(b; p)+

(w′, ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r ,−1,(w; p)

∣∣∣∣∣∣ (8.127)

≤ (b−a)M

E ,c,q,r ,−1,(w; p)

(max{| f ′(a)|q, | f ′(b)|q}) 1

q

for q < r+(), where w′ = w(b−a) .

Proof. From Lemma 8.8, properties of modulus and power mean inequality, we have∣∣∣∣∣∣f (a)+ f (b)

2−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′, ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.128)

≤ b−a

2E ,c,q,r , , (w; p)

∫ 1

0

∣∣∣(1− t)−1E ,c,q,r , , (w(1− t) ; p)− t−1E ,c,q,r

, , (wt ; p)∣∣∣

×| f ′(ta+(1− t)b|dt

≤ b−a

2E ,c,q,r , , (w; p)

(∫ 1

0|(1− t)−1E ,c,q,r

, , (w(1− t) ; p)

−t−1E ,c,q,r , , (wt ; p)|dt

)1− 1q ×

(∫ 1

0|(1− t)−1E ,c,q,r

, , (w(1− t) ; p)

−t−1E ,c,q,r , , (wt ; p)|| f ′(ta+(1− t)b|qdt

) 1q.

Since | f ′|q is quasi-convex, also using absolute convergence of Mittag-Leffler function, wehave∣∣∣∣∣∣

f (a)+ f (b)2

−(w′, ,c,q,ra+, ,−1, f

)(b; p)+

(w′, ,c,q,rb−, ,−1, f

)(a; p)

2(b−a)−1E ,c,q,r , , (w; p)

∣∣∣∣∣∣ (8.129)

≤ (b−a)M

2E ,c,q,r , , (w; p)

(∫ 1

0|(1− t)−1|dt +

∫ 1

0|t−1|dt

)(max{| f ′(a)|q, | f ′(b)|q}) 1

q .



2− ()

2(b−a)−1

[J−1a+ f (b)+ J−1

b− f (a)]∣∣∣∣≤ b−a

(max{| f ′(a)|q, | f ′(b)|q}) 1

q .

(8.130)



2− 1

b−a

∫ b

af (t)dt

∣∣∣∣≤ b−a2

(max{| f ′(a)|q, | f ′(b)|q}) 1

q .

8.6 Bounds of Fractional Integral Operatorsfor s-convex Functions

In this section bounds of fractional integral operators are proved for s-convex functions.

Theorem 8.26 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] −→ R be a real valued function. If f is positive and s-convex, then for , ′ ≥ 1,the following fractional integral inequality for generalized integral operators holds:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.131)

≤(

f (a)+ f (x)s+1

)(x−a)Hw

a+,−1(x; p)

+(

f (b)+ f (x)s+1

)(b− x)Hw

b−, ′−1(x; p),x ∈ [a,b].

Proof. As f is s-convex so for t ∈ [a,x], we have

f (t) ≤(

x− tx−a

)s

f (a)+(

t−ax−a

)s

f (x). (8.132)

First multiplying (8.48) and (8.132). Then integrating over [a,x], we get∫ x


, , (w(x− t) ; p) f (t)dt

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)(x−a)s { f (a)

∫ x

a(x− t)sdt + f (x)

∫ x

a(t−a)sdt},

and then we have(w, ,c,q,ra+, , , f

)(x; p) ≤ (x−a)Hw

a+,−1(x; p)(

f (a)+ f (x)s+1

). (8.133)

Again for t ∈ [x,b] using convexity of f we have

f (t) ≤(

t− xb− x

)s

f (b)+(

b− tb− x

)s

f (x). (8.134)

8.6 BOUNDS OF OPERATORS FOR s-CONVEX FUNCTIONS 173

Multiplying (8.52) and (8.134), then integrating over [x,b], we get

∫ b

x(t − x)

′−1E ,c,q,r , ′, (w(t − x) ; p) f (t)dt

≤(b− x)

′−1E ,c,q,r , ′, (w(b− x) ; p)

(b− x)s { f (b)∫ x

a(t− x)sdt + f (x)

∫ x

a(b− t)sdt},

and then we have(w, ,c,q,rb−, , ′, f

)(x; p) ≤ (b− x)Hw

b−, ′−1(x; p)(

f (b)+ f (x)s+1

). (8.135)

Adding (8.133) and (8.135), the required inequality (8.131) is obtained. �

Some special cases are studied in the following corollaries.

Corollary 8.20 If we set = ′ in (8.131), then we get following inequality:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , , f

)(x; p) (8.136)

≤(

f (a)+ f (x)s+1

)(x−a)Hw

a+,−1(x; p)

+(

f (b)+ f (x)s+1

)(b− x)Hw

b−,−1(x; p),x ∈ [a,b].

Corollary 8.21 Along with assumption of Theorem 1, if f ∈ L[a,b], then we get follow-ing inequality: (

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.137)

≤ || f ||s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

].

Corollary 8.22 If = ′ in (8.137), then we get following result:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , , f

)(x; p) (8.138)

≤ || f ||s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−,−1(x; p)

].

Corollary 8.23 If s = 1 in (8.137), then we get following result:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.139)

≤ || f ||2

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

].


Theorem 8.27 With the assumption of Theorem 8.26 if f ∈ L[a,b], then operators de-fined in (2.12) and (2.13) are bounded and continuous.

Proof. Let f ∈ L[a,b]. Then from (8.133) we have∣∣∣(w, ,c,q,ra+, , , f

)(x; p)

∣∣∣≤ 2|| f |||x−a|Hw−1,a+(x; p) ≤

2(b−a)Hw−1,a+(b; p)|| f ||s+1

.

(8.140)

That is ∣∣∣(w, ,c,q,ra+, , , f

)(x; p)

∣∣∣≤ M|| f ||,

where M =2(b−a)Hw

−1,a+(b;p)

s+1 .

Therefore(w, ,c,q,ra+, , , f

)(x; p) is bounded also it is easy to see that it is linear, hence this

is continuous operator. On the other hand, from (8.135) we obtain∣∣∣(w, ,c,q,rb−, , ′, f

)(x; p)

∣∣∣≤ K|| f ||,

where K =2(b−a)Hw

′−1,b− (a;p)

s+1 . Therefore(w, ,c,q,rb−, , ′, f

)(x; p) is bounded and linear, hence

it is continuous operator. �

Lemma 8.9 Let f : [a,b] → R be a s-convex function. If f is symmetric about a+b2 , then

the following inequality holds

f

(a+b

2

)≤ 1

2s−1 f (x). (8.140)

Theorem 8.28 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] −→ R, a > b, be a real valued function. If f is positive, s-convex and symmetricabout a+b

2 , then for , ′ > 0, the following fractional integral inequality for generalizedintegral operators holds:

2s−1 f

(a+b

2

)[Hw

b−, ′+1(a; p)+Hwa+,+1(b; p)

](8.141)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤[Hw

b−, ′−1(a; p)+Hwa+,−1(b; p)

](b−a)2

(f (a)+ f (b)

s+1

).

Proof. As f is s-convex so for x ∈ [a,b], we have:

f (x) ≤(

x−ab−a

)s

f (b)+(

b− xb−a

)s

f (a). (8.142)

Multiplying (8.58) and (8.142), then integrating over [a,b], we get∫ b

a(x−a)

′E ,c,q,r , ′, (w(x−a) ; p) f (x)dx

≤(b−a)

′E ,c,q,r , ′, (w(b−a) ; p)

(b−a)s

[f (b)

∫ b

a(x−a)sdx+ f (a)

∫ b

a(b− x)sdx

].


From which we have(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)

′+1E ,c,q,r , ′, (w(b−a) ; p)

(f (a)+ f (b)

s+1

)(8.143)

(w, ,c,q,rb− , , ′+1, f

)(a; p) ≤ (b−a)2Hw

b−, ′−1(a; p)(

f (a)+ f (b)s+1

). (8.144)

Multiplying (8.61) and (8.142), then integrating over [a,b], we get

∫ b

a(b− x)E ,c,q,r

, , (w(b− x) ; p) f (x)dx

≤ (b−a)E ,c,q,r , , (w(b−a) ; p)(b−a)s

[f (b)

∫ b

a(x−a)sdx+ f (a)

∫ b

a(b− x)sdx

].

From which we have(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)+1E ,c,q,r

, , (w(b−a) ; p)(

f (a)+ f (b)s+1

)(8.145)

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)2Hw

a+,−1(b; p)(

f (a)+ f (b)s+1

). (8.146)

Adding (8.144) and (8.146), we get(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p) (8.147)

≤[Hw

b−, ′−1(a; p)+Hwa+,−1(b; p)

](b−a)2

(f (a)+ f (b)

s+1

).

Multiplying (8.140) by (x−a)′E ,c,q,r , ′, (w(x−a) ; p) and integrating over [a,b]

f

(a+b

2

)∫ b

a(x−a)

′E ,c,q,r , ′, (w(x−a) ; p)dx (8.148)

≤ 12s−1

∫ b

a(x−a)

′E ,c,q,r , ′, (w(x−a) ; p) f (x)dx.

By using (2.2), we get

f

(a+b

2

)Hw

b−, ′+1(a; p) ≤ 12s−1

(w, ,c,q,rb−, , ′+1, f

)(a; p). (8.149)

Multiplying (8.140) with (b− x)E ,c,q,r , , (w(b− x) ; p) and integrating over [a,b], also

using (2.2), we get

f

(a+b

2

)Hw

a+,+1(b; p) ≤ 12s−1

(w, ,c,q,ra+, ,+1, f

)(b; p). (8.150)


By adding (8.149) and (8.150), we get;

2s−1 f

(a+b

2

)[Hw

b−, ′+1(a; p)+Hwa+,+1(b; p)

](8.151)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p).

By adding (8.147) and (8.151), inequality (8.141) is obtained. �

Corollary 8.24 If we take = ′ in (8.141), then we get

2s−1 f

(a+b

2

)[Hw

b−,+1(a; p)+Hwa+,+1(b; p)

](8.152)

≤(w, ,c,q,rb−, ,+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤[Hw

b−,−1(a; p)+Hwa+,−1(b; p)

](b−a)2

(f (a)+ f (b)

s+1

).

Theorem 8.29 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b]−→ R be a real valued function. If f is differentiable and | f ′| is s-convex, then for , ′ ≥ 1, the following fractional integral inequality for generalized integral operatorsholds: ∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)+(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.153)

−(Hw

a+,−1(x; p) f (a)+Hwb−, ′−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|+ | f ′(x)|

s+1

)(x−a)Hw

a+,−1(x; p)

+( | f ′(b)|+ | f ′(x)|

s+1

)(b− x)Hw

b−, ′−1(x; p),x ∈ [a,b].

Proof. Let x ∈ [a,b] and t ∈ [a,x], by using s-convexity of | f ′|, we have

| f ′(t)| ≤(

x− tx−a

)s

| f ′(a)|+(

t−ax−a

)s

| f ′(x)|. (8.154)

From (8.154) we have

f ′(t) ≤(

x− tx−a

)s

| f ′(a)|+(

t−ax−a

)s

| f ′(x)|. (8.155)

The product of (8.48) and (8.155), gives the following inequality

(x− t)−1E ,c,q,r , , (w(x− t) ; p) f ′(t)dt (8.156)

≤ (x−a)−2E ,c,q,r , , (w(x−a) ; p)

(| f ′(a)|(x− t)+ | f ′(x)|(t −a)).


After integrating above inequality over [a,x], we get∫ x


, , (w(x− t) ; p) f ′(t)dt (8.157)

≤ (x−a)−2E ,c,q,r , , (w(x−a) ; p){| f ′(a)|

∫ x

a(x− t)dt + | f ′(x)|

∫ x

a(t−a)dt}

= (x−a)E ,c,q,r , , (w(x− t) ; p)

( | f ′(a)|+ | f ′(x)|s+1

).

By using (8.72), (8.157) takes the following form

(Hw−1,a+(x; p)

)f (a)−

(w, ,c,q,ra+, ,+1, f

)(x; p) ≤ (x−a)Hw

−1,a+(x; p)( | f ′(a)|+ | f ′(x)|

s+1

).

(8.158)

Also, from (8.154) we get

f ′(t) ≥−((

x− tx−a

)s

| f ′(a)|+(

t−ax−a

)s

| f ′(x)|)

. (8.159)

Following the same procedure as we did for (8.155), we also have(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

−1,a+(x; p) f (a) (8.160)

≤ (x−a)Hw−1,a+(x; p)

( | f ′(a)|+ | f ′(x)|s+1

).

From (8.158) and (8.160), we get∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

−1,a+(x; p) f (a)∣∣∣ (8.161)

≤ (x−a)Hw−1,a+(x; p)

( | f ′(a)|+ | f ′(x)|s+1

).

Now we let x ∈ [a,b] and t ∈ [x,b]. Then by using s-convexity of | f ′| we have

| f ′(t)| ≤(

t− xb− x

)s

| f ′(b)|+(

b− tb− x

)s

| f ′(x)|. (8.162)

On the same lines as we have done for (8.48), (8.155) and (8.159) one can get from (8.52)and (8.162), the following inequality:∣∣∣(w, ,c,q,r

b−, , ′+1, f)

(x; p)−Hw ′−1,b−(x; p) f (b)

∣∣∣ (8.163)

≤ (b− x)Hw ′−1,b−(x; p)

( | f ′(b)|+ | f ′(x)|s+1

).

From inequalities (8.161) and (8.163) via triangular inequality (8.153) is obtained. �


Corollary 8.25 If we take = ′ in (8.153), then we get∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)+

(w, ,c,q,rb−, ,+1, f

)(x; p) (8.164)

−(Hw

a+,−1(x; p) f (a)+Hwb−,−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|+ | f ′(x)|

s+1

)(x−a)Hw

a+,−1(x; p)

+( | f ′(b)|+ | f ′(x)|

s+1

)(b− x)Hw

b−,−1(x; p),x ∈ [a,b].

8.7 Bounds of Fractional Integral Inequalitiesfor (s,m)-convex Functions

The results of this section are generalizations of results of previous section.

Theorem 8.30 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] −→ R be a real valued function. If f is positive and (s,m)-convex, then for , ′ ≥ 1, the following fractional integral inequality holds:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.165)

≤(

f (a)+mf (x)s+1

)(x−a)Hw

a+,−1(x; p)

+(

f (b)+mf (x)s+1

)(b− x)Hw

b−, ′−1(x; p),x ∈ [a,b].

Proof. Since f is (s,m)-convex, we obtain

f (t) ≤(

x− tx−a

)s

f (a)+m

(t−ax−a

)s

f (x). (8.166)

By multiplying (8.48) and (8.166) and then integrating over [a,x], we get∫ x


, , (w(x− t) ; p) f (t)dt

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)

(f (a)

(x−a)s

∫ x

a(x− t)sdt

+mf (x)∫ x

a

(t −ax−a

)s

dt

),

that is, the left integral operator satisfies the following inequality:(w, ,c,q,ra+, , , f

)(x; p) ≤ (x−a)Hw

a+,−1(x; p)(

f (a)+mf (x)s+1

). (8.167)

8.7 BOUNDS FOR (s,m)-CONVEX FUNCTIONS 179

Again from (s,m)-convexity of f , we have

f (t) ≤(

t− xb− x

)s

f (b)+m

(b− tb− x

)s

f (x). (8.168)

By multiplying (8.52) and (8.168) and then integrating over [x,b], we have

∫ b

x(t − x)

′−1E ,c,q,r , ′, (w(t − x) ; p) f (t)dt

≤ (b− x)′−1E ,c,q,r

, ′, (w(b− x) ; p)(

f (a)(b− x)s

∫ b

x(t − x)sdt

+mf (x)∫ b

x

(b− tb− x

)s

dt

),

that is, the right integral operator satisfies the following inequality:

(w, ,c,q,rb−, , ′, f

)(x; p) ≤ (b− x)Hw

b−, ′−1(x; p)(

f (b)+mf (x)s+1

). (8.169)

By Adding (8.167) and (8.169), the required inequality (8.165) is established. �

Some particular results are stated in the following corollaries.

Corollary 8.26 If we set = ′ in (8.165), then the following inequality is obtained:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , , f

)(x; p) (8.170)

≤(

f (a)+mf (x)s+1

)(x−a)Hw

a+,−1(x; p)

+(

f (b)+mf (x)s+1

)(b− x)Hw

b−,−1(x; p),x ∈ [a,b].

Corollary 8.27 Along with assumptions of Theorem 1, if f ∈ L[a,b], then the followinginequality is obtained:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.171)

≤ || f ||(1+m)s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

].

Corollary 8.28 If we take = ′ in (8.171), then we get the following result:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.172)

≤ || f ||(1+m)s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−,−1(x; p)

].


Corollary 8.29 If we take s = 1 in (8.171), then we get the following result:(w, ,c,q,ra+ , , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.173)

≤ || f ||(1+m)2

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

].

Theorem 8.31 With the assumptions of Theorem 1 if f ∈ L[a,b], then operators definedin (2.12) and (2.13) are bounded and continuous.

Proof. If f ∈ L[a,b], then from (8.167) we have

∣∣∣(w, ,c,q,ra+, , , f

)(x; p)

∣∣∣≤ 2|| f ||(1+m)|x−a|Hwa+,−1(x; p)

s+1(8.174)

≤2|| f ||(b−a)Hw

a+,−1(b; p)(1+m)

s+1,

that is ∣∣∣(w, ,c,q,ra+, , , f

)(x; p)

∣∣∣≤ M|| f ||,

where M =2(b−a)Hw

a+,−1(b;p)(1+m)

s+1 . Therefore(w, ,c,q,ra+, , , f

)(x; p) is bounded also it is easy

to see that it is linear, hence this is continuous operator. Also on the other hand from(8.169) we can obtain: ∣∣∣(w, ,c,q,r

b−, , ′, f)

(x; p)∣∣∣≤ K|| f ||,

where K =2(b−a)Hw

b−, ′−1(a;p)(1+m)

s+1 . Therefore(w, ,c,q,rb−, , ′, f

)(x; p) is bounded also it is

linear, hence continuous. �

The following lemma is needed to prove the next result.

Lemma 8.10 Let f : [a,mb]→ R be (s,m)-convex function. If f is f (a+mb− x) = f (x)and (s,m) ∈ [0,1]2, then the following inequality holds:

f

(a+mb

2

)≤ (1+m) f (x)

2s . (8.174)

Proof. As f is (s,m)-convex function, we have

f

(a+mb

2

)≤ f ((1− t)a+mtb)

2s +mf (ta+m(1− t)b)

2s . (8.175)

Let x = a(1− t)+mtb. Then we have a+mb− x = ta+m(1− t)b.

f

(a+mb

2

)≤ f (x)

2s +mf (a+mb− x)

2s . (8.176)

Hence by using f (a+mb− x) = f (x), the inequality (8.174) can be obtained. �


Theorem 8.32 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] −→ R, a > b, be a real valued function. If f is positive, (s,m)-convex andf (a+mb−x)= f (x), then for , ′ > 0, the following fractional integral inequality holds:

2s(b−a)1+m

f

(a+mb

2

)[Hw

b−, ′(a; p)+Hwa+, (b; p)

](8.177)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤[Hw

b−, ′(a; p)+Hwa+, (b; p)

](b−a)

(f (b)+mf (a)

s+1

).

Proof. As f is (s,m)-convex so for x ∈ [a,b], we have:

f (x) ≤(

x−ab−a

)s

f (b)+m

(b− xb−a

)s

f (a). (8.178)

By multiplying (8.58) and (8.178) and then integrating over [a,b], we get

∫ b

a(x−a)

′E ,c,q,r , ′+1,(w(x−a) ; p) f (x)dx

≤ (b−a)′E ,c,q,r , ′+1,(w(b−a) ; p)

(f (b)

(b−a)s

∫ b

a(x−a)sdx

+mf (a)∫ b

a

(b− xb−a

)s

dx

).

from which we have(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)

′+1E ,c,q,r , ′+1,(w(b−a) ; p)

(f (b)+ms f (a)

s+1

), (8.179)

that is (w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)Hw

b−, ′(a; p)(

f (b)+mf (a)s+1

). (8.180)

By multiplying (8.178) and (8.61)and then integrating over [a,b], we get

∫ b

a(b− x)E ,c,q,r

,+1,(w(b− x) ; p) f (x)dx

≤ (b−a)E ,c,q,r ,+1,(w(b−a) ; p)

(f (b)

(b−a)s

∫ b

a(x−a)sdx

+mf (a)∫ b

a

(b− xb−a

)s

dx

).

from which we have(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)+1E ,c,q,r

,+1,(w(b−a) ; p)(

f (b)+mf (a)s+1

), (8.181)


that is (w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)Hw

a+, (b; p)(

f (b)+ms f (a)s+1

). (8.182)

Adding (8.180) and (8.182), we get;(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p) (8.183)

≤[Hw

b−, ′(a; p)+Hwa+, (b; p)

](b−a)

(f (b)+mf (a)

s+1

).

Multiplying (8.174) with (x− a)′E ,c,q,r , ′+1,(w(x− a) ; p) and integrating over [a,b], we

get

f

(a+mb

2

)∫ b

a(x−a)

′E ,c,q,r , ′+1,(w(x−a) ; p)dx (8.184)

≤ 1+m2s

∫ b

a(x−a)

′E ,c,q,r , ′+1,(w(x−a) ; p) f (x)dx,

f

(a+mb

2

)(b−a)Hw

b−, ′(a; p) ≤ 1+m2s

(w, ,c,q,rb−, , ′+1, f

)(a; p). (8.185)

By multiplying (8.174) with (b−x)E ,c,q,r ,+1,(w(b−x) ; p) and integrating over [a,b], and

after simplification we get

f

(a+mb

2

)(b−a)Hw

a+, (b; p) ≤ 1+m2s

(w, ,c,q,ra+, ,+1, f

)(b; p). (8.186)


2s

1+mf

(a+mb

2

)(b−a)

[Hw

b−, ′(a; p)+Hwa+, (b; p)

](8.187)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p).

By combining (8.183) and (8.187), inequality (8.177) can be obtained. �

Corollary 8.30 If we take = ′ in (8.177), then the following inequality is obtained:

2s(b−a)1+m

f

(a+mb

2

)[Hw

b−, (a; p)+Hwa+, (b; p)

](8.188)

≤(w, ,c,q,rb−, ,+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤[Hw

b−, (a; p)+Hwa+, (b; p)

](b−a)

(f (b)+mf (a)

s+1

).


Theorem 8.33 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b]−→R be a real valued function. If f is differentiable and | f ′| is (s,m)-convex, thenfor , ′ ≥ 1, the following fractional integral inequality for generalized integral operatorsholds: ∣∣∣(w, ,c,q,r

a+ , ,+1, f)

(x; p)+(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.189)

−(Hw

a+,−1(x; p) f (a)+Hwb−, ′−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|+ms| f ′(x)|

s+1

)(x−a)Hw

a+,−1(x; p)

+( | f ′(b)|+ms| f ′(x)|

s+1

)(b− x)Hw

b−, ′−1(x; p),x ∈ [a,b].

Proof. Let x ∈ [a,b] and t ∈ [a,x), by using (s,m)-convexity of | f ′|, we have

| f ′(t)| ≤(

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s

| f ′(x)|. (8.190)

From (8.190) follows

f ′(t) ≤(

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s

| f ′(x)|. (8.191)

The product of (8.48) and (8.191), gives the following inequality:

(x− t)−1E ,c,q,r , , (w(x− t) ; p) f ′(t)dt (8.192)

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)

((x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s

| f ′(x)|)

.

After integrating above inequality over [a,x], we get∫ x


, , (w(x− t) ; p) f ′(t)dt (8.193)

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)

( | f ′(a)|(x−a)s

∫ x

a(x− t)sdt

+m| f ′(x)|∫ x

a

(t−ax−a

)s

dt

)

= (x−a)E ,c,q,r , , (w(x−a) ; p)

( | f ′(a)|+ms| f ′(x)|s+1

).

By using (8.72), (8.193) takes the following form:

Hwa+,−1(x; p) f (a)−

(w, ,c,q,ra+, ,+1, f

)(x; p) (8.194)

≤ (x−a)Hwa+,−1(x; p)

( | f ′(a)|+ms| f ′(x)|s+1

).


Also, from (8.190) we have

f ′(t) ≥−((

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s

| f ′(x)|)

. (8.195)

Following the same procedure as we did for (8.191), we obtain:

(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

a+,−1(x; p) f (a) (8.196)

≤ (x−a)Hwa+,−1(x; p)

( | f ′(a)|+ms| f ′(x)|s+1

).

From (8.194) and (8.196), we get

∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

a+,−1(x; p) f (a)∣∣∣ (8.197)

≤ (x−a)Hwa+,−1(x; p)

( | f ′(a)|+ms| f ′(x)|s+1

).

Now we let x ∈ [a,b] and t ∈ (x,b]. Then by using (s,m)-convexity of | f ′| we have

| f ′(t)| ≤(

t− xb− x

)s

| f ′(b)|+m

(b− tb− x

)s

| f ′(x)|. (8.198)

On the same lines as we have done for (8.48), (8.191) and (8.195) one can get from (8.52)and (8.198), the following inequality:

∣∣∣(w, ,c,q,rb−, , ′+1, f

)(x; p)−Hw

b−, ′−1(x; p) f (b)∣∣∣ (8.199)

≤ (b− x)Hwb−, ′−1(x; p)

( | f ′(b)|+ms| f ′(x)|s+1

).

From inequalities (8.197) and (8.199) via triangular inequality (8.189) is obtained. �


∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)+

(w, ,c,q,rb−, ,+1, f

)(x; p) (8.200)

−(Hw

a+,−1(x; p) f (a)+Hwb−,−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|+ms| f ′(x)|

s+1

)(x−a)Hw

a+,−1(x; p)

+( | f ′(b)|+ms| f ′(x)|

s+1

)(b− x)Hw

b−,−1(x; p),x ∈ [a,b].

8.8 BOUNDS FOR EXPONENTIALLY s-CONVEX FUNCTIONS 185

8.8 Bounds of Fractional Integral Operatorsfor Exponentially s-convex Functions

The generalizations of results proved for s-convex functions are given in this section.

Theorem 8.34 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] −→ R be a real valued function. If f is positive and exponentially s-convex, thenfor , ′ ≥ 1, the following upper bound for generalized integral operators holds:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.201)

≤(

f (a)ea +

f (x)ex

) (x−a)Hwa+,−1(x; p)

s+1

+(

f (b)eb

+f (x)e x

) (b− x)Hwb−, ′−1(x; p)

s+1,x ∈ [a,b], , ∈ R.

Proof. Since f is exponentially s-convex, we obtain

f (t) ≤(

x− tx−a

)s f (a)ea +

(t−ax−a

)s f (x)ex . (8.202)



, , (w(x− t) ; p) f (t)dt

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)(x−a)s { f (a)

ea

∫ x

a(x− t)sdt +

f (x)ex

∫ x

a(t −a)sdt},

that is, the left integral operator follow the upcoming inequality:

(w, ,c,q,ra+, , , f

)(x; p) ≤

(x−a)Hw−1,a+(x; p)

s+1

(f (a)ea +

f (x)ex

). (8.203)

Again from exponentially s-convexity of f , we have

f (t) ≤(

t − xb− x

)s f (b)eb

+(

b− tb− x

)s f (x)e x

. (8.204)

By multiplying (8.52) and (8.204) and then integrating over [x,b], we get

∫ b

x(t− x)

′−1E ,c,q,r , ′, (w(t − x) ; p) f (t)dt

≤(b− x)

′−1E ,c,q,r , ′, (w(b− x) ; p)

(b− x)s { f (b)eb

∫ b

x(t − x)sdt +

f (x)e x

∫ b

x(b− t)sdt}


that is, the right integral operator satisfies the following inequality:(w, ,c,q,rb−, , ′, f

)(x; p) ≤

(b− x)Hw ′−1,b−(x; p)

s+1

(f (b)eb

+f (x)e x

). (8.205)

By adding (8.203) and (8.205), the required inequality (8.201) can be obtained. �

Some particular results are stated in the following corollaries.


)(x; p)+

(w, ,c,q,rb−, , , f

)(x; p) (8.206)

≤(

f (a)ea +

f (x)ex

) (x−a)Hwa+,−1(x; p)

s+1

+(

f (b)eb

+f (x)e x

) (b− x)Hwb−,−1(x; p)

s+1,x ∈ [a,b].

Corollary 8.33 Along with assumption of Theorem 1, if f ∈ L[a,b], then the followinginequality is obtained:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.207)

≤ || f ||s+1

[(1

ea +1

ex

)(x−a)Hw

a+,−1(x; p)

+(

1

eb+

1

e x

)(b− x)Hw

b−, ′−1(x; p)].

Corollary 8.34 If we take = ′ in (8.207), then we get following result:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , , f

)(x; p) (8.208)

≤ || f ||s+1

[(1

ea +1

ex

)(x−a)Hw

a+,−1(x; p)

+(

1

eb+

1

e x

)(b− x)Hw

b−,−1(x; p)].

Corollary 8.35 If we take s = 1 in (8.207), then following result for exponentially convexfunctions hold: (

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.209)

≤ || f ||2

[(1

ea +1

ex

)(x−a)Hw

a+,−1(x; p)

+(

1

eb+

1

e x

)(b− x)Hw

b−, ′−1(x; p)].


Theorem 8.35 With the assumptions of Theorem 1 if f ∈ L[a,b], then operators definedin (2.12) and (2.13) are linear, bounded and continuous.

Proof. If f ∈ L[a,b], then from (8.167) we have∣∣∣(w, ,c,q,ra+, , , f

)(x; p)

∣∣∣≤ 2|| f ||(x−a)Hwa+,−1(x; p)

s+1

(1

ea +1

ex

)(8.210)

≤2(b−a)Hw

a+,−1(b; p)

s+1

(1

ea +1

ex

)|| f ||,

that is∣∣∣(w, ,c,q,r

a+, , , f)

(x; p)∣∣∣ ≤ M|| f ||, where M =

2(b−a)Hwa+,−1

(b;p)

2ea(s+1) , ≥ 0 and for

< 0, M =2(b−a)Hw

a+,−1(b;p)

s+1

(1

ea + 1eb

). Therefore

(w, ,c,q,ra+, , , f

)(x; p) is bounded also

it is easy to see that it is linear, hence this is continuous operator. On the other hand, from(8.169) we obtain: ∣∣∣(w, ,c,q,r

a+, , ′, f)

(x; p)∣∣∣≤ K|| f ||,

where K =2(b−a)Hw

b−, ′−1(a;p)

2ea(s+1), ≥ 0 and for < 0, K =

2(b−a)Hwb−, ′−1

(a;p)

s+1

(1

ea + 1eb

).

Therefore(w, ,c,q,ra+, , ′, f

)(x; p) is bounded also it is linear, hence continuous. �

Definition 8.1 Let f : [a,b]→R be a function, we say f is exponentially symmetric abouta+b2 , if

f (x)ex =

f (a+b− x)e(a+b−x) , ∈ R. (8.211)

It is required to give the following lemma which will be helpful to produce Hadamard typeestimations.

Lemma 8.11 Let f : [a,b] → R be an exponentially s-convex function. If f is exponen-tially symmetric, then the following inequality holds:

f

(a+b

2

)≤ f (x)

2s−1ex . (8.212)

Proof. For [a,b] ⊂ R be a closed interval, t ∈ [0,1] and ∈ R, we have

f(

a+b2

)= f

(at+(1−t)b

2 + a(1−t)+bt2

)Since f is exponentially s-convex, so

f

(a+b

2

)≤ f (at +(1− t)b)

2se(at+(1−t)b) +f (a(1− t)+bt)2se(a(1−t)+bt) . (8.213)

Let x = at +(1− t)b, where x ∈ [a,b]. Then we have a+b− x = bt +(1− t)a and we get

f

(a+b

2

)≤ f (x)

2se(x) +f (a+b− x)2se(a+b−x) . (8.214)

Now using the fact of exponentially symmetric we will get (8.212). �


Theorem 8.36 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] −→ R, a > b, be a real valued function. If f is positive, exponentially s-convexand symmetric about a+b

2 , then for , > 0, the following fractional integral inequality forgeneralized integral operators holds:

2s−1ex f

(a+b

2

)[Hw

b−, ′+1(a; p)+Hwa+,+1(b; p)

](8.215)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤[Hw

a+,−1(a; p)+Hwa+,−1(b; p)

] (b−a)2

s+1

(f (a)ea +

f (b)eb

).

Proof. As f is exponentially s-convex, for x ∈ [a,b], we have:

f (x) ≤(

x−ab−a

)s f (b)eb

+(

b− xb−a

)s f (a)ea . (8.216)


∫ b

a(x−a)tE ,c,q,r

, ′, (w(x−a) ; p) f (x)dx

≤ (b−a)′−sE ,c,q,r

, ′, (w(b−a) ; p)[

f (b)eb

∫ b

a(x−a)sdx+

f (a)ea

∫ b

a(b− x)sdx

].

From which we have

(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤

(b−a)′+1E ,c,q,r

, ′, (w(b−a) ; p)

s+1

(f (a)ea +

f (b)eb

), (8.217)

(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)2

s+1Hw ′−1,b−(a; p)

(f (a)ea +

f (b)eb

). (8.218)


∫ b

a(b− x)E ,c,q,r

, , (w(b− x) ; p) f (x)dx

≤ (b−a)−sE ,c,q,r , , (w(b−a) ; p)

[f (b)eb

∫ b

a(x−a)sdx+

f (a)ea

∫ b

a(b− x)sdx

].

From which we have

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)+1E ,c,q,r

, , (w(b−a) ; p)s+1

(f (a)ea +

f (b)eb

), (8.219)

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)2

s+1Hw−1,a+(b; p)

(f (a)ea +

f (b)eb

). (8.220)


Adding (8.218) and (8.220), we get;

(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p) (8.221)

≤[Hw ′−1,b−(a; p)+Hw

−1,a+(b; p)] (b−a)2

s+1

(f (a)ea +

f (b)eb

).

Multiplying (8.212) with (x−a)tE ,c,q,r , ′, (w(x−a) ; p) and integrating over [a,b], we get

f

(a+b

2

)∫ b

a(x−a)tE ,c,q,r

, ′, (w(x−a) ; p)dx (8.222)

≤ 12s−1ex

∫ b

a(x−a)tE ,c,q,r

, ′, (w(x−a) ; p) f (x)dx.

By using Definition (2.2), we get

f

(a+b

2

)Hw ′+1,b−(a; p) ≤ 1

2s−1ex

(w, ,c,q,rb−, , ′+1, f

)(a; p). (8.223)

Multiplying (8.212) with (b− x)E ,c,q,r , , (w(b− x) ; p) and integrating over [a,b], we get

f

(a+b

2

)Hw+1,a+(b; p) ≤ 1

2s−1ex

(w, ,c,q,ra+, ,+1, f

)(b; p). (8.224)


2s−1ex f

(a+b

2

)[Hw ′+1,b−(a; p)+Hw

+1,a+(b; p)]

(8.225)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p).



2s−1ex f

(a+b

2

)[Hw

b−,+1(a; p)+Hwa+,+1(b; p)

](8.226)

≤(w, ,c,q,rb−, ,+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤[Hw

b−,−1(a; p)+Hwa+,−1(b; p)

] (b−a)2

s+1

(f (a)ea +

f (b)eb

).


Theorem 8.37 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,b] −→ R, be a real valued function. If f is differentiable and | f ′| is exponentiallys-convex, then for , ′ ≥ 1, the following fractional integral inequality for generalizedintegral operators holds:∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)+(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.227)

−(Hw

a+,−1(x; p) f (a)+Hwa+,−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|

ea +| f ′(x)|ex

) (x−a)Hwa+,−1(x; p)

s+1

+( | f ′(b)|

eb+

| f ′(x)|e x

) (b− x)Hwa+,−1(x; p)

s+1,x ∈ [a,b].

Proof. Let x ∈ [a,b] and t ∈ [a,x], by using exponentially s-convexity of | f ′|, we have

| f ′(t)| ≤(

x− tx−a

)s | f ′(a)|ea +

(t−ax−a

)s | f ′(x)|ex . (8.228)

From (8.228) follows

f ′(t) ≤(

x− tx−a

)s | f ′(a)|ea +

(t−ax−a

)s | f ′(x)|ex . (8.229)

The product of (8.48) and (8.229), gives the following inequality

(x− t)−1E ,c,q,r , , (w(x− t) ; p) f ′(t)dt (8.230)

≤ (x−a)−1−sE ,c,q,r , , (w(x−a) ; p)

( | f ′(a)|ea (x− t)s +

| f ′(x)|ex (t −a)s

).

After integrating above inequality over [a,x], we get

∫ x


, , (w(x− t) ; p) f ′(t)dt (8.231)

≤ (x−a)−1−sE ,c,q,r , , (w(x−a) ; p){ | f

′(a)|ea

∫ x

a(x− t)sdt +

| f ′(x)|ex

∫ x

a(t−a)sdt}

=(x−a)E ,c,q,r

, , (w(x− t) ; p)s+1

( | f ′(a)|ea +

| f ′(x)|ex

).

By using (8.72), (8.231) takes the following form:(Hw−1,a+(x; p)

)f (a)−

(w, ,c,q,ra+, ,+1, f

)(x; p) (8.232)

≤(x−a)Hw

−1,a+(x; p)

s+1

( | f ′(a)|ea +

| f ′(x)|ex

).


Also from (8.228) we have

f ′(t) ≥−((

x− tx−a

)s | f ′(a)|ea +

(t−ax−a

)s | f ′(x)|ex

). (8.233)

Following the same procedure as we did for (8.229), one can obtain:

(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

−1,a+(x; p) f (a) (8.234)

≤(x−a)Hw

−1,a+(x; p)

s+1

( | f ′(a)|ea +

| f ′(x)|ex

).

From (8.232) and (8.234), we get

∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

−1,a+(x; p) f (a)∣∣∣ (8.235)

≤(x−a)Hw

−1,a+(x; p)

s+1

( | f ′(a)|ea +

| f ′(x)|ex

).

Now we let x ∈ [a,b] and t ∈ [x,b]. Then by exponentially s-convexity of | f ′| we have

| f ′(t)| ≤(

t− xb− x

)s | f ′(b)|eb

+(

b− tb− x

)s | f ′(x)|e x

. (8.236)

On the same lines as we have done for (8.48), (8.229) and (8.233) one can get from(8.52) and (8.236), the following inequality:

∣∣∣(w, ,c,q,rb−, , ′+1, f

)(x; p)−Hw

′−1,b−(x; p) f (b)∣∣∣ ≤ (b− x)Hw

′−1,b−(x; p)

s+1

( | f ′(b)|eb

+| f ′(x)|e x

).

(8.237)

From inequalities (8.235) and (8.237) via triangular inequality (8.227) can be obtained. �


∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)+

(w, ,c,q,rb−, ,+1, f

)(x; p) (8.238)

−(Hw

a+,−1(x; p) f (a)+Hwb−,−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|

ea +| f ′(x)|ex

) (x−a)Hwa+,−1(x; p)

s+1

+( | f ′(b)|

eb+

| f ′(x)|e x

) (b− x)Hwb−,−1(x; p)

s+1,x ∈ [a,b].


8.9 Bounds of Fractional Integral Operatorsfor Strongly (s,m)-convex Functions

The first result provides an upper bound of sum of left and right fractional integrals forstrongly (s,m)-convex functions.

Theorem 8.38 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ∈ L1[a,b],0≤ a < b. If f is a positive and strongly (s,m)-convex function on [a,mb] withmodulus > 0, m ∈ (0,1], then for , ′ ≥ 1, the following fractional integral inequalityfor generalized integral operators holds:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.239)

≤(

f (a)+mf(

xm

)s+1

−(x−ma)2

6m

)(x−a)Hw

a+,−1(x; p)

+

(f (b)+mf

(xm

)s+1

−(mb− x)2

6m

)(b− x)Hw

b−, ′−1(x; p), x ∈ [a,b].

Proof. The function f is strongly (s,m)-convex function with modulus , therefore onecan obtain

f (t) ≤(

x− tx−a

)s

f (a)+m

(t−ax−a

)s

f( x

m

)−

(x− t)(t−a)(x−ma)2

m(x−a)2 . (8.240)

By multiplying (8.48) and (8.240) and then integrating over [a,x], we get

∫ x


, , (w(x− t) ; p) f (t)dt

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)

(f (a)

(x−a)s

∫ x

a(x− t)sdt

+mf( x

m

)∫ x

a

(t−ax−a

)s

dt−(x−ma)2

m(x−a)2

∫ x

a(x− t)(t−a)dt

).

Therefore, the left fractional integral operator satisfies the following inequality:

(w, ,c,q,ra+, , , f

)(x; p) ≤ (x−a)Hw

a+,−1(x; p)

(f (a)+mf

(xm

)s+1

−(x−ma)2

6m

). (8.241)

Again, from strongly (s,m)-convexity of f , we have

f (t) ≤(

t − xb− x

)s

f (b)+m

(b− tb− x

)s

f( x

m

)−

(t− x)(b− t)(mb− x)2

m(b− x)2 . (8.242)

8.9 BOUNDS FOR STRONGLY (s,m)-CONVEX FUNCTIONS 193

By multiplying (8.52) and (8.242) and then integrating over [x,b], we have

∫ b

x(t− x)

′−1E ,c,q,r , ′, (w(t − x) ; p) f (t)dt

≤ (b− x)′−1E ,c,q,r

, ′, (w(b− x) ; p)(

f (a)(b− x)s

∫ b

x(t − x)sdt

+mf( x

m

)∫ b

x

(b− tb− x

)s

dt−(mb− x)2

m(b− x)2

∫ b

x(t− x)(b− t)dt

).

Therefore, we have that, the right integral operator satisfies the following inequality:

(w, ,c,q,rb−, , ′, f

)(x; p) ≤ (b− x)Hw

b−, ′−1(x; p)

(f (b)+mf

(xm

)s+1

−(mb− x)2

6m

). (8.243)


Some particular cases are given in the following results:

Theorem 8.39 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ∈ L1[a,b],0 ≤ a < b. If f is a positive and (s,m)-convex function on [a,mb], m ∈ (0,1],then for , ′ ≥ 1, the following fractional integral inequality for generalized integraloperators holds:

(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.244)

≤(

f (a)+mf(

xm

)s+1

)(x−a)Hw

a+,−1(x; p)

+

(f (b)+mf

(xm

)s+1

)(b− x)Hw

b−, ′−1(x; p), x ∈ [a,b].

Proof. For (s,m)-convex functions the inequality (8.241) holds as follows (by setting = 0): (

w, ,c,q,ra+, , , f

)(x; p) ≤ (x−a)Hw

a+,−1(x; p)

(f (a)+mf

(xm

)s+1

). (8.245)

Also, for (s,m)-convex functions the inequality (8.243) holds as follows (by setting = 0):

(w, ,c,q,rb−, , ′, f

)(x; p) ≤ (b− x)Hw

b−, ′−1(x; p)

(f (b)+mf

(xm

)s+1

). (8.246)

By adding (8.245) and (8.246) the inequality (8.244) can be obtained. �


Corollary 8.38 The following inequality holds for strongly convex functions by takings = m = 1 in (8.239):

(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.247)

≤(

f (a)+ f (x)2

−(x−a)2

6

)(x−a)Hw

a+,−1(x; p)

+(

f (b)+ f (x)2

−(b− x)2

6

)(b− x)J−1,b−(x; p),x ∈ [a,b].

Corollary 8.39 The following inequality holds for convex functions by taking s = m =1, = 0 in (8.239) which is proved in [36, Corollary 1]:

(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.248)

≤(

f (a)+ f (x)2

)(x−a)Hw

a+,−1(x; p)

+(

f (b)+ f (x)2

)(b− x)J−1,b−(x; p),x ∈ [a,b].

Remark 8.13 The inequality (8.247) provides refinement of inequality (8.248).

Corollary 8.40 If we set = ′ in (8.239), then the following inequality is obtained:

(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.249)

≤(

f (a)+mf(

xm

)s+1

−(x−ma)2

6m

)(x−a)Hw

a+,−1(x; p)

+

(f (b)+mf

(xm

)s+1

−(mb− x)2

6m

)(b− x)J−1,b−(x; p),x ∈ [a,b].

Corollary 8.41 If we set = ′ and = 0 in (8.239), then the following inequality isobtained for (s,m)-convex function:

(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.250)

≤(

f (a)+mf(

xm

)s+1

)(x−a)Hw

a+,−1(x; p)

+

(f (b)+mf

(xm

)s+1

)(b− x)J−1,b−(x; p),x ∈ [a,b].


Corollary 8.42 Along with assumptions of Theorem 1, if f ∈ L[a,b], then the followinginequality is obtained:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.251)

≤ || f ||(1+m)s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

]

−[

(x−ma)2(x−a)6m

Hwa+,−1(x; p)+

(mb− x)2(b− x)6m

Hwb−, ′−1(x; p)

].

Corollary 8.43 Along with assumptions of Theorem 1, if f ∈ L[a,b], = 0, then thefollowing inequality is obtained for (s,m)-convex function:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.252)

≤ || f ||(1+m)s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

]Corollary 8.44 For = ′ in (8.251), we get the following result:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.253)

≤ || f ||(1+m)s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)J−1,b−(x; p)]

−[

(x−ma)2(x−a)6m

Hwa+,−1(x; p)+

(mb− x)2(b− x)6m

J−1,b−(x; p)].

Corollary 8.45 For = ′, = 0 in (8.251), we get the following result for (s,m)-convexfunction: (

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.254)

≤ || f ||(1+m)s+1

[(x−a)Hw

a+,−1(x; p)+ (b− x)J−1,b−(x; p)]

Corollary 8.46 For s = 1 in (8.251), we get the following result:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.255)

≤ || f ||(1+m)2

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

]−[

(x−ma)2(x−a)6m

Hwa+,−1(x; p)+

(mb− x)2(b− x)6m

Hwb−, ′−1(x; p)

].


Corollary 8.47 For s = 1, = 0 in (8.251), we get the following result for (s,m)-convexfunction:

(w, ,c,q,ra+ , , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.256)

≤ || f ||(1+m)2

[(x−a)Hw

a+,−1(x; p)+ (b− x)Hwb−, ′−1(x; p)

].


Theorem 8.40 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ∈ L1[a,b],0 ≤ a < b. If f is a differentiable and | f ′| is a strongly (s,m)-convex functionon [a,mb] with modulus > 0, m ∈ (0,1], then for , ′ ≥ 1, the following fractionalintegral inequality for generalized integral operators holds:

∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)+

(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.257)

−(Hw

a+,−1(x; p) f (a)+Hwb−, ′−1(x; p) f (b)

)∣∣∣≤(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(x−ma)2

6m

)(x−a)Hw

a+,−1(x; p)

+

(| f ′(b)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(mb− x)2

6m

)(b− x)Hw

b−, ′−1(x; p), x ∈ [a,b].

Proof. As x ∈ [a,b] and t ∈ [a,x), by using strongly (s,m)-convexity of | f ′|, we have

| f ′(t)| ≤(

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s ∣∣∣ f ′( xm

)∣∣∣−(x− t)(t−a)(x−ma)2

m(x−a)2 . (8.258)

From (8.258) we get

f ′(t) ≤(

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s ∣∣∣ f ′( xm

)∣∣∣−(x− t)(t−a)(x−ma)2

m(x−a)2 . (8.259)

The product of (8.48) and (8.259), gives the following inequality:

(x− t)−1E ,c,q,r , , (w(x− t) ; p) f ′(t)dt (8.260)

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)((

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s ∣∣∣ f ′( xm

)∣∣∣−(x− t)(t−a)(x−ma)2

m(x−a)2

).


After integrating the above inequality over [a,x], we get

∫ x


, , (w(x− t) ; p) f ′(t)dt (8.261)

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)

( | f ′(a)|(x−a)s

∫ x

a(x− t)sdt

+m∣∣∣ f ′( x

m

)∣∣∣∫ x

a

(t−ax−a

)s

dt−(x−ma)2

m(x−a)2

∫ x

a(x− t)(t−a)dx

)

= (x−a)E ,c,q,r , , (w(x−a) ; p)

(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(x−ma)2

6m

).

By using (8.72), (8.261) takes the following form:

(Hw

a+,−1(x; p))

f (a)−(w, ,c,q,ra+, ,+1, f

)(x; p) (8.262)

≤ (x−a)Hwa+,−1(x; p)

(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(x−ma)2

6m

).

Also, from (8.258) follows

f ′(t) ≥−((

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s ∣∣∣ f ′( xm

)∣∣∣−(x− t)(t−a)(x−ma)2

m(x−a)2

).

(8.263)Following the same procedure as we did for (8.259), one can obtain:

(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

a+,−1(x; p) f (a) (8.264)

≤ (x−a)Hwa+,−1(x; p)

(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(x−ma)2

6m

).

From (8.262) and (8.264), we get

∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)−Hw

a+,−1(x; p) f (a)∣∣∣ (8.265)

≤ (x−a)Hwa+,−1(x; p)

(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(x−ma)2

6m

).

Now, we let x ∈ [a,b] and t ∈ (x,b]. Then by using strongly (s,m)-convexity of | f ′| wehave

| f ′(t)| ≤(

t− xb− x

)s

| f ′(b)|+m

(b− tb− x

)s ∣∣∣ f ′( xm

)∣∣∣−(t− x)(b− t)(mb− x)2

m(b− x)2 . (8.266)


Similarly, from (8.52) and (8.266) the following inequality can be obtained:∣∣∣(w, ,c,q,rb−, , ′+1, f

)(x; p)−Hw

b−, ′−1(x; p) f (b)∣∣∣ (8.267)

≤ (b− x)Hwb−, ′−1(x; p)

(| f ′(b)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(mb− x)2

6m

).

From inequalities (8.265) and (8.267) via triangular inequality, (8.257) is obtained. �

Theorem 8.41 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ∈ L1[a,b],0≤ a < b. If f is a differentiable and | f ′| is a (s,m)-convex function on [a,mb],m ∈ (0,1], then for , ′ ≥ 1, the following fractional integral inequality for generalizedintegral operators holds:∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)+(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.268)

−(Hw

a+,−1(x; p) f (a)+Hwb−, ′−1(x; p) f (b)

)∣∣∣≤(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

)(x−a)Hw

a+,−1(x; p)

+

(| f ′(b)|+m

∣∣ f ′ ( xm

)∣∣s+1

)(b− x)Hw

b−, ′−1(x; p), x ∈ [a,b].

Proof. For the (s,m)-convex function | f ′| the inequality (8.265) holds as follows (bysetting = 0): ∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)−Hwa+,−1(x; p) f (a)

∣∣∣ (8.269)

≤ (x−a)Hwa+,−1(x; p)

(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

).

Also, for the (s,m)-convex function | f ′| the inequality (8.267) holds as follows (bysetting = 0): ∣∣∣(w, ,c,q,r

b−, , ′+1, f)

(x; p)−Hwb−, ′−1(x; p) f (b)

∣∣∣ (8.270)

≤ (b− x)Hwb−, ′−1(x; p)

(| f ′(b)|+m

∣∣ f ′ ( xm

)∣∣s+1

).

By adding (8.269) and (8.270), the inequality (8.268) can be obtained. �


Corollary 8.48 If we put = ′ in (8.257), then the following inequality is obtained:∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)+

(w, ,c,q,rb−, ,+1, f

)(x; p) (8.271)


−(Hw

a+,−1(x; p) f (a)+ J−1,b−(x; p) f (b))∣∣∣

≤(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(x−ma)2

6m

)(x−a)Hw

a+,−1(x; p)

+

(| f ′(b)|+m

∣∣ f ′ ( xm

)∣∣s+1

−(mb− x)2

6m

)(b− x)J−1,b−(x; p),x ∈ [a,b].

Corollary 8.49 If we put = ′ and = 0 in (8.257), then the following inequality isobtained for (s,m)-convex function:∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)+(w, ,c,q,rb−, ,+1, f

)(x; p) (8.272)

−(Hw

a+,−1(x; p) f (a)+ J−1,b−(x; p) f (b))∣∣∣

≤(| f ′(a)|+m

∣∣ f ′ ( xm

)∣∣s+1

)(x−a)Hw

a+,−1(x; p)

+

(| f ′(b)|+m

∣∣ f ′ ( xm

)∣∣s+1

)(b− x)J−1,b−(x; p),x ∈ [a,b].

Remark 8.16 The inequality (8.271) provides refinement of the inequality (8.272).

Corollary 8.50 If we put s = m = 1 in (8.257), then the following inequality is obtainedfor strongly convex function:∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)+(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.273)

−(Hw

a+,−1(x; p) f (a)+Hwb−, ′−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|+ | f ′(x)|

2−

(x−a)2

6

)(x−a)Hw

a+,−1(x; p)

+( | f ′(b)|+ | f ′(x)|

2−

(b− x)2

6

)(b− x)Hw

b−, ′−1(x; p), x ∈ [a,b].

Corollary 8.51 If we put s = m = 1 and = 0 in (8.257), then the following inequalityis obtained for convex function which is proved in [36, Corollary 2]:∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)+(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.274)

−(Hw

a+,−1(x; p) f (a)+Hwb−, ′−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|+ | f ′(x)|

2

)(x−a)Hw

a+,−1(x; p)

+( | f ′(b)|+ | f ′(x)|

2

)(b− x)Hw

b−, ′−1(x; p), x ∈ [a,b].


The following lemma is useful to prove the next result.

Lemma 8.12 Let f : [a,mb] → R be strongly (s,m)-convex function with modulus . Iff ( a+mb−x

m ) = f (x) and (s,m) ∈ [0,1]2, m �= 0, then the following inequality holds:

f

(a+mb

2

)≤ (1+m) f (x)

2s −1

4m(a+mb− x−mx)2. (8.275)

Proof. As f is strongly (s,m)-convex function, we have

f

(a+mb

2

)≤ 1

2s

(f ((1− t)a+mtb)+mf

(ta+m(1− t)b

m

))(8.276)

− 4m

(t(1+m)(a−mb)+mb−ma)2.

Let x = a(1− t)+mtb. Then we have a+mb− x = ta+m(1− t)b.

f

(a+mb

2

)≤ f (x)

2s +mf(

a+mb−xm

)2s −

14m

(a+mb− x−mx)2. (8.277)

Hence, by using f ( a+mb−xm ) = f (x), the inequality (8.275) can be obtained. �

Theorem 8.42 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ∈ L1[a,b],0≤ a < b. If f is a positive and strongly (s,m)-convex function on [a,mb] withmodulus > 0, m ∈ (0,1] and f ( a+mb−x

m ) = f (x), then for , ′ ≥ 0, the following fractional integral inequality for generalized integral operators holds:

2s

1+m

(f

(a+mb

2

)(Hw

a+,+1(b; p)+Hwb−, ′+1(a; p))+

4m

(K1 +K2))

(8.278)

≤(w, ,c,q,ra+, ,+1, f

)(b; p)+

(w, ,c,q,rb−, , ′+1, f

)(a; p)

≤(Hw

b−, ′(a; p)+Hwa+, (b; p)

)(b−a)

(f (b)+mf

(am

)s+1

−(mb−a)2

6m

).

where

K1 = (b−a)′+2Hw

b−, ′+1(a; p)−2(1+m)(b−a)′+1Hw

b−, ′+2(a; p)+2(1+m)2Hwb−, ′+3(a; p),

K2 = (b−a)+2Hwa+, ′+1(b; p)−2(1+m)(b−a)+1Hw

a+,+2(b; p)+2(1+m)2Hwa+,+3(b; p).

Proof. As f is strongly (s,m)-convex so for x ∈ [a,b], we have:

f (x) ≤(

x−ab−a

)s

f (b)+m

(b− xb−a

)s

f( a

m

)−

(x−a)(b− x)(mb−a)2

m(b−a)2 . (8.279)


∫ b

a(x−a)E ,c,q,r

, ′+1,(w(x−a) ; p) f (x)dx


≤ (b−a)′E ,c,q,r , ′+1,(w(b−a) ; p)

(f (b)

(b−a)s

∫ b

a(x−a)sdx

+mf( a

m

)∫ b

a

(b− xb−a

)s

dx−(mb−a)2

m(b−a)2

∫ b

a(x−a)(b− x)dx

).

From which we have(w, ,c,q,rb−, , ′+1, f

)(a; p) (8.280)

≤ (b−a)′+1E ,c,q,r

, ′+1,(w(b−a) ; p)

(f (b)+mf

(am

)s+1

−(mb−a)2

6m

),

that is

(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)Hw

b−, ′(a; p)

(f (b)+mf

(am

)s+1

−(mb−a)2

6m

). (8.281)

Now, on the other hand by multiplying (8.61) and (8.279)and then integrating over [a,b],we get

∫ b

a(b− x)E ,c,q,r

,+1,(w(b− x) ; p) f (x)dx

≤ (b−a)E ,c,q,r ,+1,(w(b−a) ; p)

(f (b)

(b−a)s

∫ b

a(x−a)sdx

+mf( a

m

)∫ b

a

(b− xb−a

)s

dx−(mb−a)2

m(b−a)2

∫ b

a(x−a)(b− x)dx

).

From which we have(w, ,c,q,ra+, ,+1, f

)(b; p) (8.282)

≤ (b−a)+1E ,c,q,r ,+1,(w(b−a) ; p)

(f (b)+mf

(am

)s+1

−(mb−a)2

6m

),

that is

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)Hw

a+, (b; p)

(f (b)+mf

(am

)s+1

−(mb−a)2

6m

). (8.283)

By adding (8.281) and (8.283), the second inequality of (8.278) is obtained.To prove the first inequality; multiplying (8.275) with (x−a)E ,c,q,r

, ′+1,(w(x−a) ; p)and integrating over [a,b], we get

f

(a+mb

2

)∫ b

a(x−a)E ,c,q,r

, ′+1,(w(x−a) ; p)dx (8.284)

≤ 1+m2s

∫ b

a(x−a)E ,c,q,r

, ′+1,(w(x−a) ; p) f (x)dx


− 4m

∫ b

a(x−a)E ,c,q,r

, ′+1,(w(x−a) ; p)(a+mb− x−mx)2dx.

By using (2.13) and integrating by parts, we get

f

(a+mb

2

)Hw

b−, ′+1(a; p)

≤ 1+m2s

(w, ,c,q,rb−, , ′+1, f

)(a; p)−

4m

(w, ,c,q,rb−, , ′+1,(a+mb− x−mx)2

)(a; p).

The integral operator appearing in the last term of the right hand side is calculated asfollows:(w, ,c,q,rb−, , ′+1,(a+mb− x−mx)2

)(a; p)

=

n=0

p( +nq,c− ) ( ,c− )

(c)nq

(n+ ′+1)1

()nr

∫ b

a(x−a)

′+n(a+mb− x−mx)2dx

=

n=0

p( +nq,c− ) ( ,c− )

(c)nq

(n+ ′+1)1

()nr

((b−a)

′+n+3

′ +n+1

− 2(1+m)(b−a)′+n+3

( ′ +n+1)( ′+n+2)+

2(1+m)2(b−a)′+n+2

( ′ +n+1)( ′+n+2)( ′+n+3)

)

= (b−a)′+2Hw

b−, ′+1(a; p)−2(1+m)(b−a)′+1Hw

b−, ′+2(a; p)+2(1+m)2Hwb−, ′+3(a; p)

i.e

f

(a+mb

2

)Hw

b−, ′+1(a; p) ≤ 1+m2s

(w, ,c,q,rb−, , ′+1, f

)(a; p)−

4m((b−a)

′+2 (8.285)

×Hwb−, ′+1(a; p)−2(1+m)(b−a)

′+1Hwb−, ′+2(a; p)+2(1+m)2Hw

b−, ′+3(a; p)).

By multiplying (8.275) with (b− x)E ,c,q,r ,+1,(w(b− x) ; p) and integrating over [a,b], we

get

f

(a+mb

2

)Hw

a+, ′+1(b; p) (8.286)

≤ 1+m2s

(w, ,c,q,ra+, ,+1, f

)(b; p)−

4m

(w, ,c,q,ra+, ,+1,(a+mb− x−mx)2

)(b; p).

The integral operator appearing in last term of right hand side is calculated as follows:(w, ,c,q,ra+, ,+1,(a+mb− x−mx)2

)(b; p)

= (b−a)+2Hwa+, ′+1(b; p)−2(1+m)(b−a)+1Hw

a+,+2(b; p)+2(1+m)2Hwa+,+3(b; p).

By using it in (8.286) then adding resulting inequality in (8.285), the first inequality of(8.278) can be obtained. �


Theorem 8.43 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ∈ L1[a,b],0 ≤ a < b. If f is a positive and (s,m)-convex function on [a,mb], m ∈ (0,1]and f ( a+mb−x

m ) = f (x), then for , ′ ≥ 0, the following fractional integral inequality forgeneralized integral operators holds:

2s

1+m

(f

(a+mb

2

)(Hw

a+, ′+1(b; p)+Hwb−, ′+1(a; p))

)(8.287)

≤(w, ,c,q,ra+, ,+1, f

)(b; p)+

(w, ,c,q,rb−, , ′+1, f

)(a; p)

≤(Hw

b−, ′(a; p)+Hwa+, (b; p)

)(b−a)

(f (b)+mf

(am

)s+1

).

Proof. For the (s,m)-convex function f the inequality (8.281) holds as follows (by setting = 0):

(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)Hw

b−, ′(a; p)

(f (b)+mf

(am

)s+1

). (8.288)

Also, for the (s,m)-convex function f the inequality (8.283) holds as follows (by setting = 0):

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)Hw

a+, (b; p)

(f (b)+mf

(am

)s+1

). (8.289)

By adding (8.288) and (8.289), the second inequality in (8.287) can be obtained. For firstinequality using (8.285) for = 0 we have

f

(a+mb

2

)Hw

b−, ′+1(a; p) ≤ 1+m2s

(w, ,c,q,rb−, , ′+1, f

)(a; p). (8.290)

Also, from (8.286) for = 0 we have

f

(a+mb

2

)Hw

a+, ′+1(b; p) ≤ 1+m2s

(w, ,c,q,ra+, ,+1, f

)(b; p). (8.291)

From inequalities (8.290) and (8.291), the first inequality in (8.287) can be obtained. �


If s = m = 1 in (8.278), then the following result obtained for strongly convex function.

Corollary 8.52 Let f : [a,b] −→ R, 0 ≤ a < b, be a real valued function. If f is a posi-tive, strongly convex and f (a+b− x) = f (x), then for , ′ > 0, the following fractionalintegral inequality holds:

f

(a+b

2

)(Hw

a+, ′+1(b; p)+Hwb−, ′+1(a; p))+

4

((b−a)

′+2Hwb−, ′+1(a; p) (8.292)


−4(b−a)′+1Hw

b−, ′+2(a; p)+8Hwb−, ′+3(a; p)+ (b−a)+2

×Hwa+, ′+1(b; p)−4(b−a)+1Hw

a+,+2(b; p)+8Hwa+,+3(b; p)

)≤((

w, ,c,q,ra+ , ,+1, f

)(b; p)+

(w, ,c,q,rb−, , ′+1, f

)(a; p)

)≤(Hw

b−, ′(a; p)+Hwa+, (b; p)

)(b−a)

(f (b)+ f (a)

2−

(b−a)2

6

).

Remark 8.18 For = 0 in (8.292) we get [36, Corollary 3]. Therefore (8.292) is therefinement of [36, Corollary 3].

8.10 Bounds of Fractional Integral Operatorsfor Exponentially (s,m)-convex Functions

In this section the generalizations of results given in aforementioned sections are proved.

Theorem 8.44 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : K ⊆ [0,) −→ R be a real valued function. If f is positive and exponentially (s,m)-convex, then for a,b ∈ K,a < b and , ′ ≥ 1, the following fractional integral inequalityfor generalized integral operators holds:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.293)

≤(

f (a)ea +

mf ( xm )

exm

) (x−a)Hwa+,−1(x; p)

s+1

+

(f (b)eb

+mf ( x

m )

exm

)(b− x)Hw

b−, ′−1(x; p)

s+1,x ∈ [a,b], , ∈ R.

Proof. Since f is exponentially (s,m)-convex, we obtain

f (t) ≤(

x− tx−a

)s f (a)ea +m

(t−ax−a

)s f ( xm )

exm

, ∈ R. (8.294)



, , (w(x− t) ; p) f (t)dt

≤ (x−a)−1E ,c,q,r , , (w(x−a) ; p)(x−a)s

×(

f (a)ea

∫ x

a(x− t)sdt +

mf ( xm )

exm

∫ x

a(t−a)sdt

),

8.10 BOUNDS FOR EXPONENTIALLY (s,m)-CONVEX FUNCTIONS 205

that is, the left integral operator satisfies the following inequality:

(w, ,c,q,ra+, , , f

)(x; p) ≤

(x−a)Hw−1,a+(x; p)

s+1

(f (a)ea +m

f ( xm)

exm

). (8.295)

Again from exponentially (s,m)-convexity of f , we have

f (t) ≤(

t− xb− x

)s f (b)eb

+m

(b− tb− x

)s f ( xm )

exm

, ∈ R. (8.296)

By multiplying (8.52) and (8.296) and then integrating over [x,b], we get

∫ b

x(t− x)

′−1E ,c,q,r , ′, (w(t − x) ; p) f (t)dt

≤(b− x)

′−1E ,c,q,r , ′, (w(b− x) ; p)

(b− x)s

×(

f (b)eb

∫ b

x(t − x)sdt +

mf ( xm )

exm

∫ b

x(b− t)sdt

)

that is, the right integral operator satisfies the following inequality:

(w, ,c,q,rb−, , ′, f

)(x; p) ≤

(b− x)Hw ′−1,b−(x; p)

s+1

(f (b)eb

+mf ( x

m )

exm

). (8.297)


Some special cases are given in the following corollaries.


)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.298)

≤(

f (a)ea +

mf ( xm )

exm

) (x−a)Hwa+,−1(x; p)

s+1

+

(f (b)eb

+mf ( x

m )

exm

)(b− x)Hw

b−,−1(x; p)

s+1,x ∈ [a,b].

Corollary 8.54 Along with assumption of Theorem 1, if f ∈ L[a,b], then the followinginequality is obtained:(

w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb−, , ′, f

)(x; p) (8.299)

≤ || f ||s+1

((1

ea +m

exm

)(x−a)Hw

a+,−1(x; p)

+

(1

eb+

m

exm

)(b− x)Hw

b−, ′−1(x; p)

).


Corollary 8.55 If we take = ′ in (8.299), then we get the following result:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb− , , , f

)(x; p) (8.300)

≤ || f ||s+1

((1

ea +m

exm

)(x−a)Hw

a+,−1(x; p)

+

(1

eb+

m

exm

)(b− x)Hw

b−,−1(x; p)

).

Corollary 8.56 If we take s = 1 in (8.299), then we get the following result:(w, ,c,q,ra+, , , f

)(x; p)+

(w, ,c,q,rb− , , ′, f

)(x; p) (8.301)

≤ || f ||2

((1

ea +m

exm

)(x−a)Hw

a+,−1(x; p)

+

(1

eb+

m

exm

)(b− x)Hw

b−, ′−1(x; p)

).

Theorem 8.45 With the assumptions of Theorem 1 if f ∈ L[a,b], then operators definedin (2.12) and (2.13) are continuous.

Proof. If f ∈ L[a,b], then from (8.295) we have

∣∣∣(w, ,c,q,ra+, , , f

)(x; p)

∣∣∣≤ 2|| f ||(x−a)Hw−1,a+(x; p)

s+1

(1

ea +m

exm

)(8.302)

≤2(b−a)Hw

−1,a+(b; p)

s+1

(1

ea +m

exm

)|| f ||,

that is∣∣∣(w, ,c,q,r

a+, , , f)

(x; p)∣∣∣ ≤ M|| f ||, where M =

2(b−a)Hw−1,a+(b;p)

s+1

(1

ea + m

exm

). There-

fore(w, ,c,q,ra+, , , f

)(x; p) is bounded also it is easy to see that it is linear, hence this is con-

tinuous operator. On the other hand, from (8.297) we obtain:∣∣∣(w, ,c,q,rb−, , ′, f

)(x; p)

∣∣∣≤ K|| f ||,

where K =2(b−a)Hw

′−1,b− (a;p)

s+1

(1

ea + m

exm

). Therefore

(w, ,c,q,rb−, , ′, f

)(x; p) is bounded also

it is linear, hence continuous. �

Definition 8.2 Let f : [a,b]→R be a function, we will say f is exponentiallym-symmetricabout a+b

2 , iff (x)ex =

f ( a+b−xm )

e( a+b−xm )

, ∈ R. (8.302)


It is required to give the following lemma which will be helpful to produce Hadamard typeestimations for the generalized fractional integral operators.

Lemma 8.13 Let f : K ⊆ [0,)−→R, a,b∈K,a < mb be an exponentially (s,m)-convexfunction. If f is exponentially m-symmetric about a+b

2 , then the following inequality holds:

f

(a+b

2

)≤ (1+m)

f (x)2sex , ∈ R. (8.303)

Proof. Since f is exponentially (s,m)-convex, we have

f

(a+b

2

)≤ f (at +(1− t)b)

2se(at+(1−t)b) +mf ( a(1−t)+bt

m )

2se( a(1−t)+btm )

,t ∈ [0,1]. (8.304)

Let x = at +(1− t)b, where x ∈ [a,b]. Then we have a+b− x = bt +(1− t)a and we get

f

(a+b

2

)≤ f (x)

2sex +mf ( a+b−x

m )

2se( a+b−xm )

. (8.305)

Now using that f is exponentially m-symmetric we will get (8.303). �

Theorem 8.46 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Let f :K ⊆ [0,) −→ R, a,b ∈ K,a < b, be a real valued function. If f is positive, exponentially(s,m)-convex and exponentially m-symmetric about a+b

2 , then for , ′ > 0, the followingfractional integral inequality for generalized integral operators holds:

2sh()1+m

f

(a+b

2

)[Hw

b−, ′+1(a; p)+Hwa+,+1(b; p)

](8.306)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤[Hw

b−, ′−1(a; p)+Hwa+,−1(b; p)

] (b−a)2

s+1

(f ( a

m )

eam

+f (b)eb

),, ∈ R,

where h() = eb for < 0 and h() = ea for ≥ 0.

Proof. As f is exponentially (s,m)-convex, for x ∈ [a,b], we have:

f (x) ≤(

x−ab−a

)s f (b)eb +m

(b− xb−a

)s f ( am )

eam

, ∈ R. (8.307)


∫ b

a(x−a)E ,c,q,r

, ′, (w(x−a) ; p) f (x)dx

≤ (b−a)′−sE ,c,q,r

, ′, (w(b−a) ; p)(

f (b)eb

∫ b

a(x−a)sdx+

mf ( am )

eam

∫ b

a(b− x)sdx

),


from which we have

(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤

(b−a)′+1E ,c,q,r

, ′, (w(b−a) ; p)

s+1

(f (b)eb +

mf ( am )

eam

), (8.308)

(w, ,c,q,rb−, , ′+1, f

)(a; p) ≤ (b−a)2

s+1Hw ′−1,b−(a; p)

(f (b)eb

+mf ( a

m )

eam

). (8.309)


∫ b

a(b− x)E ,c,q,r

, , (w(b− x) ; p) f (x)dx

≤ (b−a)−sE ,c,q,r , , (w(b−a) ; p)

(f (b)eb

∫ b

a(x−a)sdx+

mf ( am )

eam

∫ b

a(b− x)sdx

).

From which we have

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)+1E ,c,q,r

, , (w(b−a) ; p)s+1

(f (b)eb

+mf ( a

m )

eam

), (8.310)

(w, ,c,q,ra+, ,+1, f

)(b; p) ≤ (b−a)2

s+1Hw−1,a+(b; p)

(f (b)eb +

mf ( am )

eam

). (8.311)

Adding (8.309) and (8.311), we get;(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p) (8.312)

≤[Hw ′−1,b−(a; p)+Hw

−1,a+(b; p)] (b−a)2

s+1

(f (b)eb +

mf ( am )

eam

).

Multiplying (8.303) with (x−a)tE ,c,q,r , ′, (w(x−a) ; p) and integrating over [a,b], we get

f

(a+b

2

)∫ b

a(x−a)tE ,c,q,r

, ′, (w(x−a) ; p)dx (8.313)

≤ m+12s

∫ b

a(x−a)tE ,c,q,r

, ′, (w(x−a) ; p)f (x)ex dx,

f

(a+b

2

)Hw ′+1,b−(a; p) ≤ m+1

2sex

(w, ,c,q,rb−, , ′+1, f

)(a; p). (8.314)

Multiplying (8.303) with (b− x)E ,c,q,r , , (w(b− x) ; p) and integrating over [a,b], we get

f

(a+b

2

)Hw+1,a+(b; p) ≤ m+1

2sh()

(w, ,c,q,ra+, ,+1, f

)(b; p). (8.315)



2sh()1+m

f

(a+b

2

)[Hw ′+1,b−(a; p)+Hw

+1,a+(b; p)]

(8.316)

≤(w, ,c,q,rb−, , ′+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p).



2sex

1+mf

(a+b

2

)[Hw

b−,+1(a; p)+Hwa+,+1(b; p)

](8.317)

≤(w, ,c,q,rb−, ,+1, f

)(a; p)+

(w, ,c,q,ra+, ,+1, f

)(b; p)

≤(Hw

b−,−1(a; p)+Hwa+,−1(b; p)

) (b−a)2

s+1

(f (b)eb

+mf ( a

m )

eam

).

Next result provides boundedness of sum of left and right integrals at an arbitrary point forfunctions whose derivatives in absolute values are exponentially (s,m)-convex.

Theorem 8.47 Let w ∈ R, r, , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : K ⊆ [0,)−→R be a real valued function. If f is differentiable and | f ′| is exponentially(s,m)-convex, then for a,b ∈ K,a < b and , ′ ≥ 1, the following fractional integralinequality for generalized integral operators holds:∣∣∣(w, ,c,q,r

a+, ,+1, f)

(x; p)+(w, ,c,q,rb−, , ′+1, f

)(x; p) (8.318)

−(Hw

a+,−1(x; p) f (a)+Hwb−, ′−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|

ea +m| f ′( x

m )|exm

) (x−a)Hwa+,−1(x; p)

s+1

+

(| f ′(b)|eb

+m| f ′( x

m)|exm

)(b− x)Hw

b−, ′−1(x; p)

s+1,x ∈ [a,b], , ∈ R.

Proof. Proof is on the same lines as the proof of Theorem 8.33. �

Corollary 8.58 If we take = ′ in (8.318), then the following inequality is obtained:∣∣∣(w, ,c,q,ra+, ,+1, f

)(x; p)+

(w, ,c,q,rb−, ,+1, f

)(x; p) (8.319)

−(Hw

a+,−1(x; p) f (a)+Hwb−,−1(x; p) f (b)

)∣∣∣≤( | f ′(a)|

ea +m| f ′( x

m )|exm

) (x−a)Hwa+,−1(x; p)

s+1

+

(| f ′(b)|eb

+m| f ′( x

m )|exm

)(b− x)Hw

b−,−1(x; p)

s+1,x ∈ [a,b],, ∈ R.


8.11 Bounds of Fractional Integral Operatorsfor Exponentially m-convex Functions

First we give the fractional Hadamard inequality for exponentially m-convex functions viageneralized fractional integral operators.

Theorem 8.48 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,mb]⊂R→R be a function such that f ∈ L1[a,mb] with a < mb. If f is exponentiallym-convex function, then the following inequalities hold:

e f( a+mb2 )Hw′

a+, (mb; p) (8.320)

≤(w′, ,c,q,ra+, , , e f

)(mb; p)+m+1

(w′m , ,c,q,rb−, , , e f

)(am ; p

)2

≤ m+1

2(mb−a)

[(e f (a)−m2e f ( a

m2 ))

Hw′m

b−,+1

( am

; p)

+(mb−a)(e f (b) +me

f ( am2 ))

Hw′m

b−,

( am

; p)]

where m ∈ (0,1] and w′ = w(mb−a) .

Proof. Since f is exponentially m-convex, we have

e f( x+my2 ) ≤ e f (x) +mef (y)

2, x,y ∈ [a,mb] and m ∈ (0,1]. (8.321)

Putting x = za+m(1− z)b and y = (1− z) am + zb in (8.321), we get

2e f( a+mb2 ) ≤ e f (za+m(1−z)b) +mef ((1−z) a

m +zb). (8.322)

Also from exponentially m-convexity of f , we have

e f (za+m(1−z)b) +mef ((1−z) am +zb) (8.323)

≤ ze f (a) +m(1− z)e f (b) +m(m(1− z)e f ( a

m2 ) + ze f (b))

= z(e f (a)−m2e

f ( am2 ))

+m(e f (b) +me

f ( am2 ))

.

Multiplying both sides of (8.334) with z−1E ,c,q,r , , (wz ; p) and integrating over [0,1], we

have

2e f( a+mb2 )

∫ 1

0z−1E ,c,q,r

, , (wz ; p)dz (8.324)

≤∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f (za+m(1−z)b)dz

8.11 BOUNDS FOR EXPONENTIALLY m-CONVEX FUNCTIONS 211

+m∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ((1−z) am+zb)dz.

Putting u = za+m(1− z)b and v = (1− z) am + zb in (8.336), we get

2e f( a+mb2 )

∫ mb

a(mb−u)−1E ,c,q,r

, , (w′(mb−u) ; p)du

≤∫ mb


, , (w′(mb−u) ; p)e f (u)du

+m+1∫ b

am

(v− a

m

)−1E ,c,q,r , ,

(mw′(v− a

m) ; p

)e f (v)dv.

From above, the first inequality of (8.320) is achieved.Now multiplying both sides of (8.335) with z−1E ,c,q,r

, , (wz ; p) and integrating over [0,1],we have ∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f (za+m(1−z)b)dz (8.325)

+m∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ((1−z) am +zb)dz

≤(e f (a)−m2e

f ( am2 ))∫ 1

0zE ,c,q,r

, , (wz ; p)dz

+m(e f (b) +mef ( a

m2 ))∫ 1

0z−1E ,c,q,r

, , (wz ; p)dz.


∫ mb


, , (w′(mb−u) ; p)e f (u)du

+m+1∫ b

am

(v− a

m

)−1E ,c,q,r , ,

(mw′(v− a

m) ; p

)e f (v)dv

≤ m+1

(mb−a)

[(e f (a)−m2e

f ( am2 ))∫ b

am

(v− a

m

)E ,c,q,r , ,

(mw′(v− a

m) ; p

)dv

+(mb−a)(e f (b) +mef ( a

m2 ))∫ b

am

(v− a

m

)−1E ,c,q,r , ,

(mw′(v− a

m) ; p

)dv

].

By using the definition of generalized integral operators, second inequality of (8.320) isachieved. �

Corollary 8.59 Suppose that assumptions of Theorem 8.48 hold and let m = 1. Thenfollowing inequalities for exponentially convex function hold:

e f( a+b2 )Hw∗

a+, (b; p) (8.326)

≤(w∗, ,c,q,ra+, , , e f

)(b; p)+

(w∗, ,c,q,rb−, , , e f

)(a; p)

2


≤ e f (a) + e f (b)

2Hw∗

b−, (a; p)

where w∗ = w(b−a) .

The following Hadamard inequality for exponentially m-convex function is proved whichhave several misprints.

Theorem 8.49 [129] Let f : [a,mb] ⊂ R → R be a function such that f ∈ L1[a,mb] witha < mb. If f is exponentially m-convex function, then the following inequalities hold:

2e f( a+mb2 )H( a+mb

2 )+, (mb; p) (8.327)

≤(w′2 , ,c,q,r

( a+mb2 )+

, , ,e f)

(mb; p)+m+1(w′2 , ,c,q,r

( a+mb2m )−, , ,

e f)( a

m; p)

≤ a(mb−a)

(e f (a)−m2e

f ( am2 ))

H( a+mb2 )+

,+1(mb; p)

+m+1(e f (b) +me

f ( am2 ))

H( a+mb2m )−, .

The correct form of the above theorem is stated and proved in the following theorem.

Theorem 8.50 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf : [a,mb]⊂R→R be a function such that f ∈ L1[a,mb] with a < mb. If f is exponentiallym-convex function, then the following inequalities hold:

e f( a+mb2 )Hw′2

( a+mb2 )+

,(mb; p) (8.328)

≤

(w′2 , ,c,q,r

( a+mb2 )+

, , ,e f

)(mb; p)+m+1

(w′(2m) , ,c,q,r

( a+mb2m )−, , ,

e f

)(am ; p

)2

≤ m+1

2(mb−a)

[(e f (a)−m2e f ( a

m2 ))

Hw′(2m)

( a+mb2m )−,+1

( am

; p)


f ( am2 ))

Hw′(2m)

( a+mb2m )−,

( am

; p)]

where m ∈ (0,1] and w′ is defined in (8.320).

Proof. Putting x = z2a+m (2−z)

2 b and y = z2b+ (2−z)

2am in (8.321), we get

2e f( a+mb2 ) ≤ e f ( z

2 a+m (2−z)2 b) +mef ( z

2 b+ (2−z)2

am ). (8.329)


have

2e f( a+mb2 )

∫ 1

0z−1E ,c,q,r

, , (wz ; p)dz (8.330)


≤∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ( z2 a+ (2−z)

2 b)dz

+m∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ( z2 b+ (2−z)

2am )dz.

Putting u = z2a+m (2−z)

2 b and v = z2b+ (2−z)


2e f( a+mb2 )

∫ mb

a+mb2

(mb−u)−1E ,c,q,r , , (2w(mb−u) ; p)du

≤∫ mb

a+mb2

(mb−u)−1E ,c,q,r , , (2w(mb−u) ; p)e f (u)du

+m+1∫ a+mb

2m

am

(v− a

m

)−1E ,c,q,r , ,

((2m)w(v− a

m) ; p

)e f (v)dv,

first inequality of (8.328) is achieved.From exponentially m-convexity of f , we have

e f ( z2 a+m (2−z)

2 b) +mef ( z2 b+ (2−z)

2am ) (8.331)

≤ z2e f (a) +m

(2− z)2

e f (b) +m

(z2e f (b) +m

(2− z)2

e f ( am2 ))

=z2

(e f (a) −m2e

f ( am2 ))

+m(e f (b) +me

f ( am2 ))

.


have ∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ( z2 a+m (2−z)

2 b)dz (8.332)

+m∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ( z2 b+ (2−z)

2am )dz

≤ 12

(e f (a) −m2e f ( a

m2 ))∫ 1

0zE ,c,q,r

, , (wz ; p)dz

+m(e f (b) +me

f ( am2 ))∫ 1

0z−1E ,c,q,r

, , (wz ; p)dz.

Putting u = z2a+m (2−z)

2 b and v = z2b+ (2−z)


∫ mb

a+mb2

(mb−u)−1E ,c,q,r , , (2w(mb−u) ; p)e f (u)du

+m+1∫ a+mb

2m

am

(v− a

m

)−1E ,c,q,r , ,

((2m)w(v− a

m) ; p

)e f (v)dv

≤ m+1

mb−a

[(e f (a)−m2e

f ( am2 ))∫ a+mb

2m

am

(v− a

m

)E ,c,q,r , ,

((2m)w(v− a

m) ; p

)dv



f ( am2 ))∫ a+mb

2m

am

(v− a

m

)−1E ,c,q,r , ,

((2m)w(v− a

m) ; p

)dv

],

second inequality of (8.328) is achieved. �


e f( a+b2 )Hw∗(2)

( a+b2 )+

,(b; p)

≤

(w∗(2) , ,c,q,r

( a+b2 )+

, , ,e f

)(b; p)+

(w∗(2) , ,c,q,r

( a+b2 )+

, , ,e f

)(a; p)

2

≤ e f (a) + e f (b)

2Hw∗(2)

( a+b2 )−,

(a; p)

where w∗ is defined in (8.59).

Remark 8.19 If we take w = p = 0 in (8.328), then [130, Theorem 3.3] is obtained.

In the following we give Fejer-Hadamard inequality for exponentially m-convex functionsvia generalized fractional integral operators.

Theorem 8.51 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r+ . Let f :[a,mb]⊂R→R be a function such that f ∈ L1[a,mb] with a < mb. Also, let g : [a,mb]→R

be a function which is nonnegative and integrable. If f is exponentially m-convex functionand f (v) = f (a+mb−mv), then the following inequalities hold:

e f( a+mb2 )

(w′m , ,c,q,rb−, , , eg

)( am

; p)

(8.333)

≤(1+m)

(w′m , ,c,q,rb−, , , e f eg

)(am ; p

)2

≤ m2(mb−a)

[(e f (a)−m2e

f ( am2 ))(

w′m , ,c,q,rb−, ,+1, eg

)( am

; p)


f ( am2 ))(

w′m , ,c,q,rb−, , , eg

)( am

; p)]

where m ∈ (0,1] and w′ is defined in (8.320).

Proof. Putting x = za+m(1− z)b and y = (1− z) am + zb in (8.321), we get

2e f( a+mb2 ) ≤ e f (za+m(1−z)b) +mef ((1−z) a

m +zb). (8.334)

Also from exponentially m-convexity of f , we have

e f (za+m(1−z)b) +mef ((1−z) am +zb) (8.335)


≤ ze f (a) +m(1− z)e f (b) +m(m(1− z)e f ( a

m2 ) + ze f (b))

= z(e f (a)−m2e

f ( am2 ))

+m(e f (b) +me

f ( am2 ))

.


have

2e f( a+mb2 )

∫ 1

0z−1E ,c,q,r

, , (wz ; p)dz (8.336)

≤∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f (za+m(1−z)b)dz

+m∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ((1−z) am+zb)dz.


2e f( a+mb2 )

∫ mb


, , (w(mb−u) ; p)du

≤∫ mb


, , (w(mb−u) ; p)e f (u)du

+m+1∫ b

am

(v− a

m

)−1E ,c,q,r , ,

(mw(v− a

m) ; p

)e f (v)dv,

first inequality of (8.320) is achieved.

Now multiplying both sides of (8.335) with z−1E ,c,q,r , , (wz ; p) and integrating over

[0,1], we have

∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f (za+m(1−z)b)dz (8.337)

+m∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ((1−z) am +zb)dz

≤(e f (a)−m2e

f ( am2 ))∫ 1

0zE ,c,q,r

, , (wz ; p)dz

+m(e f (b) +mef ( a

m2 ))∫ 1

0z−1E ,c,q,r

, , (wz ; p)dz.


∫ mb


, , (w(mb−u) ; p)e f (u)du

+m+1∫ b

am

(v− a

m

)−1E ,c,q,r , ,

(mw(v− a

m) ; p

)e f (v)dv

≤ m+1

(mb−a)

[(e f (a)−m2e f ( a

m2 ))∫ b

am

(v− a

m

)E ,c,q,r , ,

(mw(v− a

m) ; p

)dv



f ( am2 ))∫ b

am

(v− a

m

)−1E ,c,q,r , ,

(mw(v− a

m) ; p

)dv

],

second inequality of (8.320) is achieved. �


e f( a+b2 )

(w∗, ,c,q,rb−, , , eg

)(a; p) ≤

(w∗, ,c,q,rb−, , , e f eg

)(a; p)

≤ e f (a) + e f (b)

2

(w∗, ,c,q,rb−, , , eg

)(a; p)


Theorem 8.52 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,g : [a,mb] ⊂ R → R be the functions such that f ,g ∈ L1[a,mb] with a < mb. If f and gare exponentially m-convex functions, then the following inequality holds:(

w∗, ,c,q,rmb−, , ,e

f)

(a; p)+(w′, ,c,q,ra+, , , eg

)(mb; p) (8.338)

≤ 1(mb−a)

[(eg(a) +mef (b)

)(w′, ,c,q,ra+, ,+1,

)(mb; p)

+(e f (a) +meg(b)

){(mb−a)Hw′

a+, (mb; p)−Hw′a+,+1 (mb; p)

}]where m ∈ (0,1] and w′ is defined in (8.320).

Proof. Since f and g are exponentially m-convex, we have

e f ((1−z)a+mzb) + eg(za+m(1−z)b) ≤ (1− z)(e f (a) +meg(b)

)+ z

(eg(a) +mef (b)

). (8.339)


have ∫ 1

0z−1E ,c,q,r

, , (wz ; p)e f ((1−z)a+mzb)dz (8.340)

+∫ 1

0z−1E ,c,q,r

, , (wz ; p)eg(za+m(1−z)b)dz

≤(e f (a) +meg(b)

)∫ 1

0(1− z)z−1E ,c,q,r

, , (wz ; p)dz

+(eg(a) +mef (b)

)∫ 1

0zE ,c,q,r

, , (wz ; p)dz.

Putting u = (1−z)a+mzb and v = za+m(1−z)b in (8.340), inequality (8.338) is achieved.�


Corollary 8.62 Suppose that assumptions of Theorem 8.52 hold and let m = 1. Thenfollowing inequality for exponentially convex function holds:(

w∗, ,c,q,rb−, , , eg

)(a; p)+

(w∗, ,c,q,ra+, , , eg

)(b; p)

≤ 1(b−a)

[(eg(a) + e f (b)

)Hw∗

a+,+1 (b; p)

+(e f (a) + eg(b)

){(b−a)Hw∗

a+, (b; p)−Hw∗a+,+1 (b; p)

}].


Remark 8.20 If we take w = p = 0 in (8.338), then [130, Theorem 3.2] is obtained.

Further generalizations of above proved results are given in the forthcoming results.

Theorem 8.53 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,h : [a,mb] ⊂ R → R, 0 < a < mb be the real valued-functions. If f be a integrableand exponentially m-convex and h be a differentiable and strictly increasing. Then thefollowing inequalities hold:

2ef(

h(a)+mh(b)2

) (h

w′, ,c,q,ra+, , , 1

)(h−1(mh(b)); p) (8.341)

≤(

hw′, ,c,q,ra+, , , e f◦h

)(h−1(mh(b)); p)+m+1

(h

w′m , ,c,q,rb−, , , e f◦h

)(h−1

(h(a)m

); p

)

≤ m+1

(mh(b)−h(a))

[(e f (h(a))−m2e

f(

h(a)m2

))(h

w′m , ,c,q,rb−, , , 1

)(h−1

(h(a)m

); p

)

+(

e f (h(b)) +mef(

h(a)m2

))(mh(b)−h(a))

(h

w′m , ,c,q,rb−, , , 1

)(h−1

(h(a)m

); p

)]

where w′ = w(mh(b)−h(a)) .

Proof. Since f is exponentially m-convex function on [a,mb], for t ∈ [0,1], we have

2ef(

h(a)+mh(b)2

)≤ e f (th(a)+m(1−t)h(b)) +mef ((1−t) h(a)

m +th(b)). (8.342)

Also, from exponentially m-convexity, we have

e f (th(a)+m(1−t)h(b)) +mef ((1−t) h(a)m +th(b)) (8.343)

≤ t

(e f (h(a))−m2e

f(

h(a)m2

))+m

(e f (h(b)) +me

f(

h(a)m2

)).

Multiplying both sides of (8.342) with t−1E ,c,q,r , , (wt ; p) and integrating over [0,1], we

have

2ef(

h(a)+mh(b)2

) ∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt (8.344)


≤∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f (th(a)+m(1−t)h(b))dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f ((1−t) h(a)m +th(b))dt.

Putting h(u) = th(a)+m(1− t)h(b) and h(v) = (1− t) h(a)m + th(b) in (8.344), we get

2ef(

h(a)+mh(b)2

) ∫ h−1(mh(b))

a(mh(b)−h(u))−1E ,c,q,r

, , (w′(mh(b)−h(u)) ; p)d(h(u))

≤∫ h−1(mh(b))

a(mh(b)−h(u))−1E ,c,q,r

, , (w′(mh(b)−h(u)) ; p)e f (h(u))d(h(u))

+m+1∫ b

h−1(

h(a)m

)(

h(v)− h(a)m

)−1

E ,c,q,r , ,

(mw′

(h(v)− h(a)

m

); p

)e f (h(v))d(h(v)).

By using the definitions of involved fractional integral operators, the first inequality of(8.341) is obtained.

Now multiplying both sides of (8.343) with t−1E ,c,q,r , , (wt ; p) and integrating over

[0,1], we have ∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f (th(a)+m(1−t)h(b))dt (8.345)

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f ((1−t) h(a)m +th(b))dt

≤(

e f (h(a))−m2ef(

h(a)m2

))∫ 1

0tE ,c,q,r

, , (wt ; p)dt

+m

(e f (h(b)) +me

f(

h(a)m2

))∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt.

Putting h(u)= th(a)+m(1−t)h(b) and h(v) = (1−t) h(a)m +th(b) in (8.345), then by using

the definition of involved fractional integral operators, the second inequality of (8.341) isobtained. �

Corollary 8.63 Under the assumptions of Theorem 8.53 if we take m = 1, then we getfollowing inequalities for exponentially convex function:

2ef(

h(a)+h(b)2

) (h

w′, ,c,q,ra+, , , 1

)(b; p) (8.346)

≤(

hw′, ,c,q,ra+, , , e f◦h

)(b; p)+

(h

w′ , ,c,q,rb−, , , e f◦h

)(a; p)

≤(e f (h(b)) + e f (h(a))

)(h

w′, ,c,q,rb−, , , 1

)(a; p)

where w′ = w(h(b)−h(a)) .


Remark 8.21

(i) If h(u) = u in (8.341), then Theorem 8.48 is obtained.

(ii) If h(u) = u and m = 1 in (8.341), then [87, Corollary 2.2] is obtained.

(iii) If h(u) = u in (8.346), then [87, Corollary 2.2] is obtained.

In the following we give another version of the Hadamard inequality for generalizedfractional integral operators.

Theorem 8.54 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,h : [a,mb] ⊂ R → R, 0 < a < mb be the real-valued functions. If f be a integrableand exponentially m-convex and h be a differentiable and strictly increasing. Then thefollowing inequalities hold:

2ef(

h(a)+mh(b)2

)(h

w′2 , ,c,q,r(h−1

(h(a)+mh(b)

2

))+, , ,

1

)(h−1(mh(b)); p) (8.347)

≤(

hw′2 , ,c,q,r(h−1

(h(a)+mh(b)

2

))+, , ,

e f◦h)

(h−1(mh(b)); p)

+m+1

(h

w′(2m) , ,c,q,r(h−1

(h(a)+mh(b)

2m

))−, , ,

e f◦h)(

h−1(

h(a)m

); p

)

≤ m+1

(mh(b)−h(a))

[(e f (h(a))−m2e

f(

h(a)m2

))(h

w′(2m) , ,c,q,r(h−1

(h(a)+mh(b)

2m

))−, ,+1,

1

)(h−1

(h(a)m

); p

)

+(e f (h(b)) +me

f(

h(a)m2

))(mh(b)−h(a))

(h

w′(2m) , ,c,q,r(h−1

(h(a)+mh(b)

2m

))−, , ,

1

)(h−1

(h(a)m

); p

)],


Proof. Since f is exponentially m-convex function on [a,mb], for t ∈ [0,1], we have

2ef(

h(a)+mh(b)2

)≤ e

f(

t2 h(a)+m (2−t)

2 h(b))+me

f(

t2 h(b)+ (2−t)

2h(a)m

). (8.348)

Also, from exponentially m-convexity, we have

ef(

t2 h(a)+m (2−t)

2 h(b))+me

f(

t2 h(b)+ (2−t)

2h(a)m

)(8.349)

≤ t2

(e f (h(a))−m2e

f(

h(a)m2

))+m

(e f (h(b)) +me

f(

h(a)m2

)).

Multiplying both sides of (8.348) with t−1E ,c,q,r , , (wt ; p) and integrating over [0,1], we

have

2ef(

h(a)+mh(b)2

) ∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt (8.350)


≤∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(a)+m (2−t)

2 h(b))dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(b)+ (2−t)

2h(a)m

)dt.

Putting h(u) = t2h(a)+m (2−t)

2 h(b) and h(v) = t2h(b)+ (2−t)

2h(a)m in (8.350), we get

2ef(

h(a)+mh(b)2

)∫ h−1(mh(b))

h−1(

h(a)+mh(b)2

)(mh(b)−h(u))−1E ,c,q,r , , (2w(mh(b)−h(u)) ; p)d(h(u))

≤∫ h−1(mh(b))

h−1(

h(a)+mh(b)2

)(mh(b)−h(u))−1E ,c,q,r , , (2w(mh(b)−h(u)) ; p)e f (h(u))d(h(u))

+m+1∫ h−1

(h(a)+mh(b)

2m

)h−1

(h(a)m

)(h(v)− h(a)

m

)−1

E ,c,q,r , ,

((2m)w(h(v)− h(a)

m) ; p

)e f (h(v))d(h(v)).

By using (2.4), the first inequality of (8.347) is obtained. Now multiplying both sides of(8.349) with t−1E ,c,q,r

, , (wt ; p) and integrating over [0,1], we have

∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(a)+m (2−t)

2 h(b))dt (8.351)

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(b)+ (2−t)

2h(a)m

)dt

≤ 12

(e f (h(a))−m2e

f(

h(a)m2

))∫ 1

0tE ,c,q,r

, , (wt ; p)dt

+m

(e f (h(b)) +me

f(

h(a)m2

))∫ 1

0t−1E ,c,q,r

, , (wt ; p)dt.

Putting h(u) = t2h(a)+m (2−t)

2 h(b) and h(v) = t2h(b)+ (2−t)

2h(a)m in (8.351), then by using

(2.4), the second inequality of (8.347) is obtained. �


2ef(

h(a)+h(b)2

)(h

w′(2) , ,c,q,r(h−1

(h(a)+h(b)

2

))+, , ,

1

)(b; p) (8.352)

≤(

hw′(2) , ,c,q,r(h−1

(h(a)+h(b)

2

))+, , ,

e f◦h)

(b; p)+

(h

w′(2) , ,c,q,r(h−1

(h(a)+h(b)

2

))−, , ,

e f◦h)

(a; p)

≤(e f (h(b)) + e f (h(a))

)(h

w′(2) , ,c,q,r(h−1

(h(a)+h(b)

2

))−, , ,

1

)(a; p) ,



Remark 8.22




Theorem 8.55 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,h : [a,mb] ⊂ R → R, 0 < a < mb be the real-valued functions. If f be a integrable,exponentially m-convex and f (h(v)) = f (h(a)+mh(b)−mh(v)) and h be a differentiableand strictly increasing. Also, let : [a,mb] → R be a function which is nonnegative andintegrable. Then the following inequalities hold:

2ef(

h(a)+mh(b)2

) (h

w′m , ,c,q,rb−, , , e◦h

)(h−1

(h(a)m

); p

)(8.353)

≤ (1+m)(

hw′m , ,c,q,rb−, , , e f◦he◦h

)(h−1

(h(a)m

); p

)

≤ m(mh(b)−h(a))

[(e f (h(a))−m2e

f(

h(a)m2

))(h

w′m , ,c,q,rb−, ,+1, e◦h

)(h−1

(h(a)m

); p

)

+(

e f (h(b)) +mef(

h(a)m2

))(mh(b)−h(a))

(h

w′m , ,c,q,rb− , , , e◦h

)(h−1

(h(a)m

); p

)],


Proof. Multiplying both sides of (8.342) with t−1E ,c,q,r , , (wt ; p)e((1−t) h(a)

m +h(b)) andintegrating over [0,1], we have

2ef(

h(a)+mh(b)2

) ∫ 1

0t−1E ,c,q,r

, , (wt ; p)e((1−t) h(a)m +h(b))dt (8.354)

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f (h(a)+m(1−t)h(b))e((1−t) h(a)m +h(b))dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f ((1−t) h(a)m +h(b))e((1−t) h(a)

m +h(b))dt.

Putting h(v) = (1− t) h(a)m + h(b) in (8.354), we get

2ef(

h(a)+mh(b)2

)∫ b

h−1(

h(a)m

)(

h(v)− h(a)m

)−1

E ,c,q,r , ,

(mw(h(v)− h(a)

m) ; p

)e(h(v))d(h(v))

≤∫ b

h−1(

h(a)m

)(

h(v)−h(a)m

)−1

E ,c,q,r , ,

(mw(h(v)−h(a)

m) ; p

)e f (h(a)+mh(b)−mh(v))e(h(v))d(h(v))

+m∫ b

h−1(

h(a)m

)(

h(v)− h(a)m

)−1

E ,c,q,r , ,

(mw(h(v)− h(a)

m) ; p

)e f (h(v))e(h(v))d(h(v)).


By using (2.4) and given condition f (h(v)) = f (h(a)+mh(b)−mh(v)), the first inequalityof (8.353) is obtained.

Now multiplying both sides of (8.343) with t−1E ,c,q,r , , (wt ; p)e((1−t) h(a)

m +h(b)) and inte-grating over [0,1], we have

∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f (h(a)+m(1−t)h(b))e((1−t) h(a)m +h(b))dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f ((1−t) h(a)m +h(b))e((1−t) h(a)

m +h(b))dt.

≤(

e f (h(a))−m2ef(

h(a)m2

))∫ 1

0tE ,c,q,r

, , (wt ; p)e((1−t) h(a)m +h(b))dt

+m

(e f (h(b)) +me

f(

h(a)m2

))∫ 1

0t−1E ,c,q,r

, , (wt ; p)e((1−t) h(a)m +h(b))dt.

From above the second inequality of (8.353) is achieved. �


2ef(

h(a)+h(b)2

) (h

w′, ,c,q,rb−, , , e◦h

)(a; p) ≤ 2

(h

w′, ,c,q,rb−, , , e f◦he◦h

)(a; p) (8.355)

≤(e f (h(b)) + e f (h(a))

)(h

w′, ,c,q,rb−, , , e◦h

)(a; p) ,


Remark 8.23




In the following we give another generalized fractional version of the Fejer-Hadamardinequality.

Theorem 8.56 Let w ∈ R, r, , , > 0, c > > 0 with p ≥ 0 and 0 < q ≤ r + . Letf ,h : [a,mb] ⊂ R → R, 0 < a < mb be the real-valued functions. If f be a integrable,exponentially m-convex and f (h(v)) = f (h(a)+mh(b)−mh(v)) and h be a differentiableand strictly increasing. Also, let : [a,mb] → R be a function which is nonnegative andintegrable. Then the following inequalities hold:

2ef(

h(a)+mh(b)2

)(h

w′(2m) , ,c,q,r(h−1

(h(a)+mh(b)

2m

))−, , ,

e◦h)(

h−1(

h(a)m

); p

)(8.356)

≤ (1+m)

(h

w′(2m) , ,c,q,r(h−1

(h(a)+mh(b)

2m

))−, , ,

e f◦he◦h)(

h−1(

h(a)m

); p

)


≤ m(mh(b)−h(a))

[(e f (h(a))−m2e

f(

h(a)m2

))(

hw′(2m) , ,c,q,r(h−1

(h(a)+mh(b)

2m

))−, , ,

e◦h)(

h−1(

h(a)m

); p

)

+(

e f (h(b)) +mef(

h(a)m2

))(mh(b)−h(a))(

hw′(2m) , ,c,q,r(h−1

(h(a)+mh(b)

2m

))−, , ,

e◦h)(

h−1(

h(a)m

); p

)],


Proof. Multiplying both sides of (8.348) with t−1E ,c,q,r , , (wt ; p)e

(t2 h(b)+ (2−t)

2h(a)m

)and

integrating over [0,1], we have

2ef(

h(a)+mh(b)2

) ∫ 1

0t−1E ,c,q,r

, , (wt ; p)e(

t2 h(b)+ (2−t)

2h(a)m

)dt (8.357)

≤∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(a)+m (2−t)

2 h(b))e(

t2 h(b)+ (2−t)

2h(a)m

)dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(b)+ (2−t)

2h(a)m

)e(

t2 h(b)+ (2−t)

2h(a)m

)dt.

Putting h(v) = t2h(b)+ (2−t)

2h(a)m in (8.357), we get

2ef(

h(a)+mh(b)2

) ∫ (h−1

(h(a)+mh(b)

2m

))h−1

(h(a)m

)(

h(v)− h(a)m

)−1

×E ,c,q,r , ,

((2m)w(h(v)− h(a)

m) ; p

)e(h(v))d(h(v))

≤∫ (

h−1(

h(a)+mh(b)2m

))h−1

(h(a)m

)(

h(v)− h(a)m

)−1

E ,c,q,r , ,

((2m)w(h(v)− h(a)

m) ; p

)

× e f (h(a)+mh(b)−mh(v))e(h(v))d(h(v))+m∫ (

h−1(

h(a)+mh(b)2m

))h−1

(h(a)m

)(

h(v)− h(a)m

)−1

×E ,c,q,r , ,

((2m)w(h(v)− h(a)

m) ; p

)e f (h(v))e(h(v))d(h(v)).

By using (2.4) and given condition f (h(v)) = f (h(a)+mh(b)−mh(v)), the first inequalityof (8.356) is obtained.

Now multiplying both sides of (8.349) with t−1E ,c,q,r , , (wt ; p)e

(t2 h(b)+ (2−t)

2h(a)m

)and in-

tegrating over [0,1], we have


∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(a)+m (2−t)

2 h(b))e(

t2 h(b)+ (2−t)

2h(a)m

)dt

+m∫ 1

0t−1E ,c,q,r

, , (wt ; p)e f(

t2 h(b)+ (2−t)

2h(a)m

)e(

t2 h(b)+ (2−t)

2h(a)m

)dt.

≤(

e f (h(a))−m2ef(

h(a)m2

))∫ 1

0tE ,c,q,r

, , (wt ; p)e(

t2 h(b)+ (2−t)

2h(a)m

)dt

+m

(e f (h(b)) +me

f(

h(a)m2

))∫ 1

0t−1E ,c,q,r

, , (wt ; p)e(

t2 h(b)+ (2−t)

2h(a)m

)dt.

From above the second inequality of (8.356) is achieved. �


2ef(

h(a)+h(b)2

)(h

w′2 , ,c,q,r(h−1

(h(a)+h(b)

2

))−, , ,

e◦h)

(a; p) (8.358)

≤ 2

(h

w′2 , ,c,q,r(h−1

(h(a)+h(b)

2

))−, , ,

e◦h)

(a; p)

≤(e f (h(b)) + e f (h(a))

)(h

w′2 , ,c,q,r(h−1

(h(a)+h(b)

2

))−, , ,

e◦h)

(a; p) ,


Chapter9Bounds of Unified IntegralOperators ContainingMittag-Leffler Function

In this chapter bonds of unified integral operators are given for different kinds of convexfunctions. The results are further deducible for various types of fractional integral opera-tors.

This chapter is based on our results from [56, 98, 72, 109].

9.1 Bounds of Unified Integral Operatorsof Convex Functions

In this section bounds bonds of unified integral operators are given for convex functions.

Theorem 9.1 Let f : [a,b]→R be a positive convex function, 0 < a < b and g : [a,b]→R

be differentiable and strictly increasing function. Also let x be an increasing function on

[a,b] and c, ,,r ∈ C, p, ,c ≥ 0, and 0 < q ≤ c+ . Then for x ∈ [a,b] we have(gF

, ,c,q,ra+, , , f

)(x,w; p) ≤ E ,c,q,r

, , (w(g(x)−g(a)) ; p)((g(x)−g(a)))( f (x)+ f (a))

(9.1)

225

226 9 BOUNDS OF UNIFIED INTEGRAL OPERATORS

and(gF

, ,c,q,rb−, , , f

)(x,w; p) ≤ E ,c,q,r

, , (w(g(b)−g(x)) ; p)((g(b)−g(x)))( f (x)+ f (b))

(9.2)

hence (gF

, ,c,q,ra+, , , f

)(x,w; p)+

(gF

, ,c,q,rb−, , , f

)(x,w; p) (9.3)

≤ E ,c,q,r , , (w(g(x)−g(a)) ; p)((g(x)−g(a)))

( f (x)+ f (a))+E ,c,q,r , , (w(g(b)−g(x)) ; p)((g(b)−g(x)))( f (x)+ f (b)) .

Proof. The function g is increasing, therefore for t ∈ [a,x], x ∈ (a,b), g(x) − g(t)≤ g(x)−g(a). The function

x is increasing, therefore one can obtain:

(g(x)−g(t))g(x)−g(t)

≤ (g(x)−g(a))g(x)−g(a)

. (9.4)

Now by multiplying with E ,c,q,r , , (w(g(x)−g(t)) ; p)g′(t) the following inequality is yielded:


g′(t)E ,c,q,r , , (w(g(x)−g(t)) ; p)

≤ (g(x)−g(a))g(x)−g(a)

g′(t)E ,c,q,r , , (w(g(x)−g(t)) ; p). (9.5)

Also E ,c,q,r , , (w(g(x)−g(t)) ; p) is series of positive terms, therefore

E ,c,q,r , , (w(g(x)−g(t)) ; p)≤ E ,c,q,r

, , (w(g(x)−g(a)) ; p) so the following inequality holds:


g′(t)E ,c,q,r , , (w(g(x)−g(t)) ; p)

≤ (g(x)−g(a))g(x)−g(a)

g′(t)E ,c,q,r , , (w(g(x)−g(a)) ; p). (9.6)

Using convexity of f on [a,x] for x ∈ (a,b) we have

f (t) ≤ x− tx−a

f (a)+t−ax−a

f (x). (9.7)

Multiplying (9.6) and (9.7), then integrating with respect to t over [a,x] we have

∫ x

a


g′(t) f (t)E ,c,q,r , , (w(g(x)−g(t)) ; p)dt

≤ f (a)x−a

(g(x)−g(a))g(x)−g(a)

E ,c,q,r , , (w(g(x)−g(a)) ; p)

∫ x

a(x− t)g′(t)dt

+f (x)x−a


E ,c,q,r , , (w(g(x)−g(a)) ; p)

∫ x

a(t−a)g′(t)dt. (9.8)

9.1 BOUNDS OF UNIFIED INTEGRAL OPERATORS OF CONVEX FUNCTIONS 227

By using (2.21), and integrating by parts we get(gF

, ,c,q,ra+, , , f

)(x,w; p) ≤ E ,c,q,r

, , (w(g(x)−g(a)) ; p)

×((g(x)−g(a))

g(x)−g(a)

)(f (a)x−a

(g(a)(a− x)+

∫ x

ag(t)dt

)

+f (x)x−a

((x−a)g(x)−

∫ x

ag(t)dt

))

which further simplifies as follows:(gF

, ,c,q,ra+, , , f

)(x,w; p) ≤ E ,c,q,r

, , (w(g(x)−g(a)) ; p)((g(x)−g(a)))( f (x)+ f (a)) .

(9.9)

Now on the other hand for t ∈ (x,b], x ∈ (a,b) the following inequality holds true:

(g(t)−g(x))g(t)−g(x)

g′(t)E ,c,q,r , , (w(g(t)−g(x)) ; p)

≤ (g(b)−g(x))g(b)−g(x)

g′(t)E ,c,q,r , , (w(g(t)−g(x)) ; p). (9.10)

Also E ,c,q,r , , (w(g(t)−g(x)) ; p) is series of positive terms, therefore

E ,c,q,r , , (w(g(t)− g(x)) ; p) ≤ E ,c,q,r

, , (w(g(b)− g(x)) ; p), so the following inequality isvalid:


g′(t)E ,c,q,r , , (w(g(t)−g(x)) ; p)

≤ (g(b)−g(x))g(b)−g(x)

g′(t)E ,c,q,r , , (w(g(b)−g(x)) ; p). (9.11)

The following inequality also holds for convex function f :

f (t) ≤ t− xb− x

f (b)+b− tb− x

f (x). (9.12)

Multiplying (9.11) and (9.12), then integrating with respect to t over (x,b] and adoptingthe same pattern of simplification as we did for (9.8), the following inequality is obtained:(

gF , ,c,q,rb−, , , f

)(x,w; p) ≤ E ,c,q,r

, , (w(g(b)−g(x)) ; p)

×((g(b)−g(x))

g(b)−g(x)

)(f (b)b− x

(g(b)(b− x)−

∫ b

xg(t)dt

)

+f (x)b− x

((x−b)g(x)+

∫ b

xg(t)dt

))

which further simplifies as follows:(gF

, ,c,q,rb−, , , f

)(x,w; p) ≤ E ,c,q,r

, , (w(g(b)−g(x)) ; p)((g(b)−g(x)))( f (x)+ f (b)) .

(9.13)


By adding (9.9) and (9.13), (9.3) can be achieved.Henceforth we give consequences of above theorem for fractional calculus and integraloperators defined in [47, 34, 81, 92, 108, 139]. �

Proposition 9.1 Let (t) = t and p = w = 0. Then (2.21) and (2.22) produce the thefollowing bound for fractional integral operators defined in [92], as follows:

(g Ja+ f )(x)+ (g Jb− f )(x) ≤ 1()

((g(x)−g(a))

( f (x)+ f (a))+ (g(b)−g(x))( f (x)+ f (b))) ≥ 1.

Proposition 9.2 Let g(x) = I(x) = x and p = w = 0. Then (2.21) and (2.22) produce thefollowing bound for fractional integral operators defined in [138] as follows:

(a+J f )(x)+ (b−J f )(x)≤ (x−a)( f (x)+ f (a))+(b− x)( f (x)+ f (b)).

Corollary 9.1 If we take (t) = ()tk

kk() and p = w = 0. Then (2.21) and (2.22) producethe following bound for the fractional integral operators defined in [97] as follows:

(g Jka+ f )(x)+ (g Jk

b− f )(x)

≤ 1kk()

((g(x)−g(a))k ( f (x)+ f (a))

+ (g(b)−g(x))k ( f (b)+ f (x)).

Corollary 9.2 If we take (t) = t and g(x) = I(x) = x with p = w = 0. Then (2.21) and(2.22) produce the following bound for left and right Riemann-Liouville fractional integraldefined in [92] as follows:

(Ja+ f )(x)+ (Jb− f )(x)

≤ 1()

((x−a) ( f (x)+ f (a))+ (b− x)( f (b)+ f (x))) ≥ k.

Corollary 9.3 If we take (t) = tk ()

kk() and g(x) = I(x) = x, p = w = 0. Then (2.21) and(2.22) produce the following bound for the fractional integral operators define in [108] asfollows:

(Jkb− f )(x)+ (Jk

b− f )(x)

≤ 1kk()

((x−a)k ( f (x)+ f (a))

+ (b− x)k ( f (b)+ f (x))) ≥ k.


Corollary 9.4 If we take (t) = t , > 0 and g(x) = x , > 0 with p = w = 0. Then

(2.21) and (2.22) produce following bound for the fractional integral operators defined in[34], as follows:

(Ja+ f )(x)+ (Jb− f )(x)

≤ 1()

((x −a) ( f (x)+ f (a))

+ (b − x) ( f (b)+ f (x))).

Corollary 9.5 If we take (t) = t , > 0 and g(x) = xs+1

s+1 , s > 0, p = w = 0. Then (2.21)and (2.22) produce following bound for the fractional integral operators define as follows:

(sJa+ f )(x)+ (sJb− f )(x)

≤ 1(s+1)()

((xs+1 −as+1) ( f (x)+ f (a))

+ (bs+1− xs+1) ( f (b)+ f (x))).

Corollary 9.6 If we take (t) = tk ()

kk() and g(x) = xs+1

s+1 , s > 0, p = w = 0. Then (2.21)and (2.22) produce following bound for the fractional integral operators defined in [139],as follows:

(skJ

a+ f )(x)+ (skJ

b− f )(x)

≤ 1

(s+1)k kk()

(( f (x)+ f (a))(bs+1− xs+1)k

+(xs+1−as+1)k ( f (b)+ f (x))) ≥ k.

Corollary 9.7 If we take (t) = t and g(x) = x+s

+s , ,s > 0, p = w = 0. Then (2.21) and(2.22) produce following bound for the fractional integral operators defined as follows:

(s Ja+ f )(x)+ (s Jb− f )(x)

≤ 1( + s)()

((x+s−a+s) ( f (x)+ f (a))

+ (b+s− x+s) ( f (b)+ f (x))).

Corollary 9.8 If we take g(x) = (x−a) , > 0 in (2.21) and g(x) = −(b−x)

, > 0 in(2.22) with (t) = t , > 0, p = w = 0. Then (2.21) and (2.22) produce the fractionalintegral operators defined in [81], as follows:

(Ja+ f )(x) =1−

()

∫ x

a((x−a) − (t−a))−1(t−a)−1 f (t)dt

and

(Jb− f )(x) =1−

()

∫ b

x((b− x) − (b− t))−1(b− t)−1 f (t)dt.


Further they satisfy the following bound:

(Ja+ f )(x)+ (Jb− f )(x) ≤ 1()

((x−a)( f (x)+ f (a))+ (b− x)( f (b)+ f (x))).

Corollary 9.9 If we take g(x) = (x−a) , > 0 in (2.21) and g(x) = −(b−x)

, > 0 in

(2.22) with (t) = tk ()

kk() , > k, p = w = 0. Then (2.21) and (2.22) produce the fractionalintegral operators defined as follows:

(k Ja+ f )(x)+ (k Jb− f )(x)

≤ 1

k kk()

((x−a)k ( f (x)+ f (a))+ (b− x)

k ( f (b)+ f (x))).

Lemma 9.1 [47] Let f : [a,b] → R be a convex function. If f is symmetric abouta+b

2,

then the following inequality holds:

f

(a+b

2

)≤ f (x), x ∈ [a,b]. (9.14)

The following theorem provides the Hadamard type estimation of integral operators (2.21)and (2.22).

Theorem 9.2 Let f : [a,b]→R be a positive convex function, 0 < a < b and g : [a,b]→R

be differentiable and strictly increasing function. Also let x be an increasing function on

[a,b] and c, ,,r ∈ C, p, ,c ≥ 0, and 0 < q ≤ c+ , if in addition f is symmetric abouta+b

2. Then for x ∈ [a,b] we have

f

(a+b

2

)((gF

, ,c,q,rb−, , , 1

)(a; p)+ (

(gF

, ,c,q,ra+, , , 1

)(b; p)

)≤((

gF , ,c,q,rb−, , , f

)(a,w; p)+

(gF

, ,c,q,ra+, , , f

)(b,w; p)

)≤ 2(g(b)−g(a))E ,c,q,r

, , (w(g(b)−g(a)) ; p)( f (a)+ f (b)). (9.15)

Proof. For x ∈ (a,b), under the assumption on g andx

the following inequality holds:


g′(x)E ,c,q,r , , (w(g(x)−g(a)) ; p) (9.16)

≤ (g(b)−g(a))g(b)−g(a)

g′(x)E ,c,q,r , , (w(g(b)−g(a)) ; p).

Using convexity of f on [a,b] for x ∈ (a,b) we have

f (x) ≤ x−ab−a

f (b)+b− xb−a

f (a). (9.17)


Multiplying (9.16) and (9.17) and then integrating with respect to x over [a,b], the follow-ing inequality is obtained:

∫ b

a


E ,c,q,r , , (w(g(x)−g(a)) ; p)g′(x) f (x)dx

≤ f (b)b−a

(g(b)−g(a))g(b)−g(a)

E ,c,q,r , , (w(g(b)−g(a)) ; p)

∫ b

a(x−a)g′(x)dx

+f (a)b−a

(g(b)−g(a))g(b)−g(a)

E ,c,q,r , , (w(g(b)−g(a)) ; p)

∫ b

a(b− x)g′(x)dx.

By using (2.21) and integrating by parts we get(gF

, ,c,q,ra+, , , f

)(b,w; p) ≤ E ,c,q,r

, , (w(g(b)−g(a)) ; p)((g(b)−g(a)))( f (a)+ f (b)) .

(9.18)

On the other hand the following inequality holds:

(g(b)−g(x))g(b)−g(x)

g′(x)E ,c,q,r , , (w(g(b)−g(x)) ; p)

≤ (g(b)−g(a))g(b)−g(a)

g′(x)E ,c,q,r , , (w(g(b)−g(a)) ; p). (9.19)

Multiplying (9.17) and (9.19) and then integrating with respect to x over [a,b] and simpli-fying on the same pattern as we did for (9.16) and (9.17), following inequality is obtained:(

gF , ,c,q,rb−, , , f

)(a,w; p) ≤ E ,c,q,r

, , (w(g(b)−g(a)) ; p)((g(b)−g(a)))( f (a)+ f (b)) .

(9.20)

By adding (9.18) ) and (9.20), we have(gF

, ,c,q,ra+, , , f

)(b,w; p)+

(gF

, ,c,q,rb−, , , f

)(a,w; p)

≤ 2(g(b)−g(a))E ,c,q,r , , (w(g(b)−g(a)) ; p)( f (a)+ f (b)). (9.21)

Multiplying both sides of 9.14 by(g(x)−g(a))

g(x)−g(a)g′(x)E ,c,q,r

, , (w(g(x)− g(a)) ; p), then

integrating over [a,b] we get

f

(a+b

2

)∫ b

a


g′(x)E ,c,q,r , , (w(g(x)−g(a)) ; p)dx

≤∫ b

a


g′(x) f (x)E ,c,q,r , , (w(g(x)−g(a)) ; p) f (x)dx

By using 2.22 we get

f

(a+b

2

)(gF

, ,c,q,rb−, , , 1

)(a; p) ≤

(gF

, ,c,q,rb−, , , f

)(a,w; p) (9.22)


Multiplying both sides of 9.14 by(g(b)−g(x))

g(b)−g(x)g′(x)E ,c,q,r

, , (w(g(b)−g(x)) ; p) and in-

tegrating over [a,b] we have

f

(a+b

2

)(gF

, ,c,q,ra+, , , 1

)(b; p) ≤

(gF

, ,c,q,ra+, , , f

)(b,w; p) (9.23)

by adding 9.22 and 9.23, the following inequality is obtained:

f

(a+b

2

)((gF

, ,c,q,rb−, , , 1

)(a; p)+

(gF

, ,c,q,ra+, , , 1

)(b; p)

).

≤((

gF , ,c,q,rb−, , , f

)(a,w; p)+

(gF

, ,c,q,ra+, , , f

)(b,w; p)

). (9.24)

Combining 9.21 and 9.24, inequality 9.15 can be achieved. �

Theorem 9.3 Let f : [a,b] → R be a differentiable function. If | f ′| is convex, 0 < a < band g : [a,b] → R be differentiable and strictly increasing function. Also let

x be anincreasing function and c, ,,r ∈ C, p, ,c ≥ 0, and 0 < q ≤ c+ . Then for x ∈ (a,b)we have ∣∣∣(gF

, ,c,q,ra+, , , f ∗ g

)(x,w; p)+

(gF

, ,c,q,rb−, , , f ∗ g

)(x,w; p)

∣∣∣≤ E ,c,q,r

, , (w(g(x)−g(a)) ; p)(g(x)−g(a))(| f ′(x)|+ | f ′(a)|)+E ,c,q,r , , (w(g(b)−g(x)) ; p)(g(b)−g(x))(| f ′(x)|+ | f ′(b)|). (9.25)

Where(gF

, ,c,q,ra+, , , f ∗ g

)(x,w; p) :=

∫ x

a


E ,c,q,r , , (w(g(x)−g(t)) ; p)g′(t) f ′(t)dt

(gF

, ,c,q,rb−, , , f ∗ g

)(x,w; p) :=

∫ b

x


E ,c,q,r , , (w(g(x)−g(t)) ; p)g′(t) f ′(t)dt.

Proof. Using the convexity of | f ′| over [a,b] for t ∈ [a,x] we have

| f ′(t)| ≤ x− tx−a

| f ′(a)|+ t−ax−a

| f ′(x)|. (9.26)

From which we can write

−(

x− tx−a

| f ′(a)|+ t −ax−a

| f ′(x)|)≤ f ′(t) ≤

(x− tx−a

| f ′(a)|+ t−ax−a

| f ′(x)|)

(9.27)

we consider the right hand side inequality of the above inequality i.e.

f ′(t) ≤(

x− tx−a

| f ′(a)|+ t−ax−a

| f ′(x)|)

. (9.28)


Further the following inequality holds true:


g′(x)E ,c,q,r , , (w(g(x)−g(t)) ; p)

≤ (g(x)−g(a))g(x)−g(a)

g′(t)E ,c,q,r , , (w(g(x)−g(a)) ; p). (9.29)

Multiplying (9.28) and (9.29) and integrating with respect to t over [a,x], the followinginequality is obtained:

∫ x

a


g′(t) f ′(t)E ,c,q,r , , (w(g(x)−g(t)) ; p)dt

≤ | f ′(a)|x−a


E ,c,q,r , , (w(g(x)−g(a)) ; p)

∫ x

a(x− t)g′(t)dt

+| f ′(x)|x−a


E ,c,q,r , , (w(g(x)−g(a)) ; p)

∫ x

a(t −a)g′(t)dt

which gives (gF

, ,c,q,ra+, , , f ∗ g

)(x,w; p) ≤ E ,c,q,r

, , (w(g(x)−g(a)) ; p)

×(g(x)−g(a))(| f ′(x)|+ | f ′(a)|). (9.30)

If we consider the left hand side inequality from the inequality (9.27) and proceed as wedid for the right hand side inequality we have(

gF , ,c,q,ra+, , , f ∗ g

)(x,w; p) ≥−E ,c,q,r

, , (w(g(x)−g(a)) ; p)

×(g(x)−g(a))(| f ′(x)|+ | f ′(a)|) (9.31)

�

Combining (9.30) and (9.31), the following inequality is obtained:∣∣∣(gF , ,c,q,ra+, , , f ∗ g

)(x,w; p)

∣∣∣≤ E ,c,q,r , , (w(g(x)−g(a)) ; p)

×(g(x)−g(a))(| f ′(x)|+ | f ′(a)|). (9.32)

On the other hand using convexity of | f ′(t)| over [a,b] for t ∈ (x,b] we have

| f ′(t)| ≤ t− xb− x

| f ′(b)|+ b− tb− x

| f ′(x)|. (9.33)

Further the following inequality holds true:


g′(x)E ,c,q,r , , (w(g(x)−g(t)) ; p)

≤ (g(b)−g(x))g(b)−g(x)

g′(t)E ,c,q,r , , (w(g(x)−g(a)) ; p). (9.34)


By adopting the same treatment as we did for (9.26) and (9.29), one can obtain the follow-ing inequality from (9.33) and (9.34):∣∣∣(gF

, ,c,q,rb−, , , f ∗ g

)(x,w; p)

∣∣∣≤ E ,c,q,r , , (w(g(b)−g(x)) ; p)

×(g(b)−g(x))(| f ′(x)|+ | f ′(b)|). (9.35)

Combining (9.32) and (9.35), inequality (9.25) can be achieved.

9.2 Bounds of Unified Integral Operatorsfor Exponentially (s,m)-Convex Functions

Theorem 9.4 Let f : [a,b] → R, 0 ≤ a < b, be a positive exponentially (s,m)-convexfunction with m ∈ (0,1] and g : [a,b] → R be a differentiable and strictly increasing func-

tion. Also letx

be an increasing function on [a,b]. If ,, ,c ∈ C, p, ≥ 0, c ≥ 0 and

0 < q ≤ c+ , then for x ∈ (a,b) we have(gF

, ,c,q,ra+, , , f

)(x,w; p)+

(gF

, ,c,q,rb−, , , f

)(x,w; p) (9.36)

≤ Kax (E ,c,q,r

, , ,g;)

((m

f(

xm

)e( x

m ) g(x)− f (a)ea g(a)

)

−(s+1)(x−a)s

(m

f(

xm

)e( x

m )sJx−g(a)− f (a)

easJa+g(a)

))

+Kxb(E

,c,q,r , , ,g;)

((f (b)eb

g(b)−mf(

xm

)e( x

m ) g(x)

)

−(s+1)(b− x)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).

Proof. Using exponentially (s,m)-convexity of f , we have

f (t) ≤(

x− tx−a

)s f (a)ea +m

(t−ax−a

)s f(

xm

)e( x

m ) . (9.37)

Multiplying (9.6) and (9.37) and integrating over [a,x], we can obtain:

∫ x

aKt

x(E ,c,q,r , , ,g;) f (t)d(g(t)) ≤ f (a)

ea Kax (E ,c,q,r

, , ,g;)∫ x

a

(x− tx−a

)s

d(g(t)) (9.38)

+mf(

xm

)e( x

m ) Kax (E ,c,q,r

, , ,g;)∫ x

a

(t−ax−a

)s

d(g(t)).


By using Definition 2.3 and integrating by part, the following inequality is obtained:

(gF

, ,c,q,ra+, , , f

)(x,w; p) ≤ Ka

x (E ,c,q,r , , ,g;)

((m

f(

xm

)e( x


)(9.39)

−(s+1)(x−a)s

(m

f(

xm

)e( x


easJa+g(a)

)).

Using exponentially (s,m)-convexity of f , we have

f (t) ≤(

t− xb− x

)s f (b)eb +m

(b− tb− x

)s f(

xm

)e( x

m ) . (9.40)

Adopting the same pattern as we did for (9.6) and (9.37), we obtained the following in-equality from (9.11) and (9.40):

(gF

, ,c,q,rb−, , , f

)(x,w; p) ≤ Kx

b(E ,c,q,r , , ,g;)

((f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)(9.41)

−(s+1)(b− x)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).

By adding (9.39) and (9.41), (9.36) can be obtained. �

Remark 9.1

(i) If we take p = w = = 0 in (9.36), then we get [96, Theorem 1].

(ii) If we take (s,m) = (1,1) and = 0 in (9.36), then we get [98, Theorem 1].

(iii) If we take (s,m) = (1,1) and p = w = 0 in (9.36), then we get [155, Theorem 1].

(iv) If we take (s,m) = (1,1), p = w = 0 and = 0 in (9.36), then we get [104, Theorem1].

(v) If we take (t) = t , > 0, = 0, p = w = 0 and (s,m) =(1,1) in (9.36), then weget [59, Corollary 1].

(vi) If we take (t) = t , > 0, g(x) = x, = 0, p = w = 0 and (s,m) =(1,1) in (9.36),then we get [47, Corollary 1].

Proposition 9.3 Let (t) = t , and p = w = 0. Then (9.36) gives the following boundfor ≥ 1 defined in [92]:

(g Ja+ f )(x)+ (g Jb− f )(x)

≤ (g(x)−g(a))−1

()

((m

f(

xm

)e( x


)

−(s+1)(x−a)s

(m

f(

xm

)e( x


easJa+g(x)

))


+(g(b)−g(x))−1

()

((f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)

−(s+1)(b− x)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).

Proposition 9.4 Let g(x) = I(x) = x and p = w = 0. Then (9.36) gives the followingbound defined in [138]:

(a+J f )(x)+ (b−J f )(x)

≤ (x−a))(x−a)s+1

((x−a)s

(m

f(

xm

)e( x


)

−(s+1)

(m

f(

xm

)e( x


easJa+g(x)

))

+(b− x))(b− x)s+1

((b− x)s

(f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)

−(s+1)

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).


kk() and p = w = 0, then (9.36) gives the followingbound holds for ≥ k defined in [97]:

(g Jka+ f )(x)+ (g Jk

b− f )(x)

≤ (g(x)−g(a))k −1

kk()

((m

f(

xm

)e( x


)

−(s+1)(x−a)s

(m

f(

xm

)e( x


easJa+g(x)

))

+(g(b)−g(x))

k −1

kk()

((f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)

−(s+1)(b− x)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).

Corollary 9.11 If we take (t) = t , p = w = 0 and g(x) = I(x) = x, then (9.36) givesthe following bound for ≥ 1 defined in [92]:

(Ja+ f )(x)+ (Jb− f )(x)

≤ (x−a)−1

()

((m

f(

xm

)e( x


)

−(s+1)(x−a)s

(m

f(

xm

)e( x


easJa+g(x)

))


+(b− x)−1

()

((f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)

−(s+1)(b− x)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).


kk() , p = w = 0 and g(x) = I(x) = x, then (9.36)gives the following bound for ≥ k defined in [108]:

(Jka+ f )(x)+ (Jk

b− f )(x)

≤ (x−a))k −1

kk()

((m

f(

xm

)e( x


)

−(s+1)(x−a)s

(m

f(

xm

)e( x


easJa+g(x)

))

+(b− x))

s−1

kk()

((f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)

−(s+1)(b− x)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).

Corollary 9.13 If we take (t) = t , p = w = 0 and g(x) = x , > 0, then (9.36) gives

the following bound defined in [34]:

(Ja+ f )(x)+ (Jb− f )(x)

≤ (x −a)−1

()−1

((m

f(

xm

)e( x


)

−(s+1)(x−a)s

(m

f(

xm

)e( x


easJa+g(x)

))

+(b − x)−1

()−1

((f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)

−(s+1)(x−a)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).

Corollary 9.14 If we take (t) = t , p = w = 0 and g(x) = xn+1

n+1 , n > 0, then (9.36) givesthe following bound:

(nJa+ f )(x)+ (nJb− f )(x)

≤ (xn+1−an+1)−1

()(n+1)−1

((m

f(

xm

)e( x


)

−(s+1)(x−a)s

(m

f(

xm

)e( x


easJa+g(x)

))


+(bn+1− xn+1)−1

()(n+1)−1

((f (b)eb g(b)−m

f(

xm

)e( x

m ) g(x)

)

−(s+1)(x−a)s

(f (b)eb

sJb−g(x)−mf(

xm

)e( x

m )sJx+g(b)

)).

The following lemma is essential to prove the next result.

Lemma 9.2 Let f : [0,]→R, be an exponentially (s,m)-convex function with m∈ (0,1].

If 0 ≤ a < b andf (x)ex =

f ( a+b−xm )

e( a+b−xm )

, then the following inequality holds:

f

(a+b

2

)≤ 1

2s (1+m)f (x)ex , x ∈ [a,b]. (9.42)

Theorem 9.5 Under the assumptions of Theorem 9.4, in addition iff (x)ex =

f ( a+b−xm )

e( a+b−xm )

,

then we have

h()2s f

(a+b

2

)(1+m)

((gF

, ,c,q,rb−, , , 1

)(a,w; p)+

(gF

, ,c,q,ra+, , , 1

)(b,w; p)

)(9.43)

≤(

gF , ,c,q,rb−, , , f

)(a,w; p)+

(gF

, ,c,q,ra+, , , f

)(b,w; p)

≤ 2Kab (E ,c,q,r

, , ,g;)

((f (b)eb g(b)−m

f(

am

)e( a

m ) g(a)

)

−(s+1)(b−a)s

(f (b)eb

sJb−g(a)−mf(

am

)e( a

m )sJa+g(b)

)).

Proof. Using exponentially (s,m)-convexity of f , we have

f (x) ≤(

x−ab−a

)s f (b)eb +m

(b− xb−a

)s f(

am

)e( a

m ) . (9.44)

Multiplying (9.16) and (9.44) and integrating the resulting inequality over [a,b], we canobtain obtain:

∫ b

aKa

x (E ,c,q,r , , ,g;)d(g(x))

≤ mf(

am

)e( a

m ) Kab (E ,c,q,r

, , ,g;)∫ b

a

(b− xb−a

)s

d(g(x))

+f (b)eb Ka

b (E ,c,q,r , , ,g;)

∫ b

a

(x−ab−a

)s

d(g(x)).


By using Definition 2.3 and integrating by parts, the following inequality is obtained:(gF

, ,c,q,ra+, , , f

)(b,w; p) ≤ Ka

b (E ,c,q,r , , ,g;)

((f (b)eb g(b)−m

f(

am

)e( a

m ) g(a))

(b−a)s (9.45)

− (s+1)(b−a)s

(f (b)eb

sJb−g(a)−mf(

am

)e( a

m )sJa+g(b)

)).

Adopting the same pattern of simplification as we did for (9.16) and (9.44), the followinginequality can be observed for (9.44) and (9.19)(

gF , ,c,q,rb−, , , f

)(a,w; p) ≤ Ka

b (E ,c,q,r , , ,g;)

((f (b)eb g(b)−m

f(

am

)e( a

m ) g(a))

(9.46)

− (s+1)(b−a)s

(f (b)eb

sJb−g(a)−mf(

am

)e( a

m )sJa+g(b)

)).

By adding (9.45) and (9.46), following inequality can be obtained:(gF

, ,c,q,ra+, , , f

)(b,w; p)+

(gF

, ,c,q,rb−, , , f

)(a,w; p) (9.47)

≤ Kab (E ,c,q,r

, , ,g;)((

f (b)eb g(b)−m

f(

am

)e( a

m ) g(a))

− ( +1)(b−a)s

(f (b)eb

sJb−g(b)−mf(

am

)e( a

m )sJa+g(b)

)).

Multiplying both sides of (9.42) by Kax (E ,c,q,r

, , ,g;)d(g(x)), and integrating over [a,b] wehave

f

(a+b

2

)∫ b

aKa

x (E ,c,q,r , , ,g;)d(g(x))

≤(

12s

)(1+m)

∫ b

aKa

x (E ,c,q,r , , ,g;)

f (x)ex d(g(x)).

From Definition 2.3, the following inequality is obtained:

h() f

(a+b

2

)2s

(1+m)

(gF

, ,c,q,rb−, , , 1

)(a,w; p) ≤

(gF

, ,c,q,rb−, , , f

)(a,w; p). (9.48)

Similarly multiplying both sides of (9.42) by Kxb(E

,c,q,r , , ,g;)d(g(x)), and integrating over

[a,b] we have

h() f

(a+b

2

)2s

(1+m)

(gF

,, ,k,c, ,l,a+ 1

)(b,w; p) ≤

(gF

,, ,k,c, ,l,a+ f

)(b,w; p) (9.49)

by adding (9.48) and (9.49) following inequality is obtained:

h() f

(a+b

2

)2s

(1+m)

((gF

, ,c,q,rb−, , , 1

)(a,w; p)+

(gF

,, ,k,c, ,l,a+ 1

)(b,w; p)

)(9.50)

≤(

gF , ,c,q,rb−, , , f

)(a,w; p)+

(gF

,, ,k,c, ,l,a+ f

)(b,w; p).

Using (9.47) and (9.50), inequality (9.43) can be achieved. �


Remark 9.2

(i) If (s,m) = (1,1) and = 0 in (9.43), then we get [98, Theorem 2].

(ii) If (s,m) = (1,1) and p = w = 0 in (9.43), then we get [155, Theorem 4].

(iii) If p = w = = 0 in (9.43), then we get [96, Theorem 3].

(iv) If (s,m) = (1,1) and p = w = = 0 in (9.43), then we get [104, Theorem 3].


kk() and p = w = 0, then the inequality (9.43) pro-duces the following Hadamard inequality defined in [97]:

h()2s+1 f

(a+b

2

)k()(m+1)

(g(b)−g(a))/k ≤(g Jk

b− f (a)+g Jk

a+ f (b))

(9.51)

≤ 2(g(b)−g(a))k −1

kk()

((f (b)eb

g(b)−mf(

am

)e( a

m ) g(a))

− (s+1)(b−a)s

(f (b)eb

sJb−g(a)−mf(

am

)e( a

m )sJa+g(b)

)).

Corollary 9.16 If we take (t) = t and p = w = 0, then the inequality (9.43) producesthe following Hadamard inequality defined in [92]:

h()2s+1 f

(a+b

2

)()(m+1)

(g(b)−g(b)) ≤ (g Jb− f (a)+

g Ja+ f (b))

(9.52)

≤ 2(g(b)−g(a))−1

()

((f (b)eb g(b)−m

f(

am

)e( a

m ) g(a))

− (s+1)(b−a)s

(f (b)eb

sJb−g(a)−mf(

am

)e( a

m )sJa+g(b)

)).


kk() , p = w = 0 and g as identity function, then theinequality (9.43) produces the following Hadamard inequality defined in [108]:

h()2s+1 f

(a+b

2

)k()(m+1)

(b−a)/k ≤(Jk

b− f (a)+ Jka+ f (b)

)(9.53)

≤ 2(b−a)k −1

kk()

((f (b)eb g(b)−m

f(

am

)e( a

m ) g(a))

− (s+1)(b−a)s

(f (b)eb

sJb−g(a)−mf(

am

)e( a

m )sJa+g(b)

)).


Corollary 9.18 If we take (t) = t , p = w = 0 and g as identity function then the in-equality (9.43) produces the following Hadamard inequality defined in [92]:

h()2s+1 f

(a+b

2

)()(m+1)

(g(b)−g(a)) ≤ (Jb− f (a)+ Ja+ f (b))

(9.54)

≤ 2(b−a)−1

()

((f (b)eb g(b)−m

f(

am

)e( a

m ) g(a))

− (s+1)(b−a)s

(f (b)eb

sJb−g(a)−mf(

am

)e( a

m )sJa+g(b)

)).

Theorem 9.6 Let f : I → R be a differentiable function. If | f ′| is exponentially (s,m)-convex with m ∈ (0,1] and g : I → R be a differentiable and strictly increasing function.Also let

x be an increasing function on I then for a,b ∈ I, a < b. If ,, ,c ∈ C, p, ≥0, c ≥ 0 and 0 < q ≤ c+ , then for x ∈ (a,b) we have∣∣∣(gF

, ,c,q,ra+, , , f ∗ g

)(x,w; p)+

(gF

, ,c,q,rb−, , , f ∗ g

)∣∣∣(x,w; p) (9.55)

≤ Kax (E ,c,q,r

, , ,g;)

((m| f ′ ( x

m

) |e( x

m ) g(x)− | f ′(a)|ea g(a)

)

−(s+1)(x−a)s

(m| f ′ ( x

m

) |e( x

m )sJx−g(a)− | f ′(a)|

easJa+g(x)

))

+Kxb(E

,c,q,r , , ,g;)

((| f ′(b)|eb g(b)−m

| f ′( xm

) |e( x

m ) g(x)

)

− (s+1)(x−a)s

(| f ′(b)|eb

sJb−g(x)−m| f ′ ( x

m

) |e( x

m )sJx+g(b)

)),

where (gF

, ,c,q,ra+, , , f ∗ g

)(x,w; p) :=

∫ x

aKt

x(E ,c,q,r , , ,g;) f ′(t)d(g(t)),

(gF

, ,c,q,rb−, , , f ∗ g

)(x,w; p) :=

∫ b

xKx

t (E ,c,q,r , , ,g;) f ′(t)d(g(t)).

Proof. Using exponentially (s,m)-convexity of | f ′| we have

| f ′(t)| ≤(

x− tx−a

)s | f ′(a)|ea +m

(t−ax−a

)s | f ′ ( xm

) |e( x

m ) . (9.56)

(9.56) can be written as follows:

−((

x− tx−a

)s | f ′(a)|ea +m

(t−ax−a

)s | f ′ ( xm

) |e( x

m )

)≤ f ′(t) (9.57)

≤(

x− tx−a

)s | f ′(a)|ea +m

(t−ax−a

)s | f ′ ( xm

) |e( x

m ) .


Let we consider the second inequality of (9.57)

f ′(t) ≤(

x− tx−a

)s | f ′(a)|ea +m

(t−ax−a

)s | f ′ ( xm

) |e( x

m ) . (9.58)

Multiplying (9.6) and (9.58) and integrating over [a,x], we can obtain:∫ x

aKt

x(E ,c,q,r , , ,g;) f (t)d(g(t))

≤ | f ′(a)|ea Ka

x (E ,c,q,r , , ,g;)

∫ x

a

(x− tx−a

)s

d(g(t))+m| f ′ ( x

m

) |e( x

m )

×Kax (E ,c,q,r

, , ,g;)∫ x

a

(t −ax−a

)s

d(g(t)).


, ,c,q,ra+, , , f

)(x,w; p) ≤ Ka

x (E ,c,q,r , , ,g;) (9.59)

×((

m| f ′ ( x

m

) |e( x

m ) g(x)− | f ′(a)|ea g(a)

)

− (s+1)(x−a)s

(m| f ′ ( x

m

) |e( x

m )sJx−g(a)− | f ′(a)|

easJa+g(x)

)).

Now we consider the left hand side from the inequality (9.57) and adopting the same pat-tern as we did for the right hand side inequality we have(

gF , ,c,q,ra+, , , f

)(x,w; p) ≥−Ka

x (E ,c,q,r , , ,g;) (9.60)

×((

m| f ′ ( x

m

) |e( x

m ) g(x)− | f ′(a)|ea g(a)

)

− (s+1)(x−a)s

(m| f ′ ( x

m

) |e( x

m )sJx−g(a)− | f ′(a)|

easJa+g(x)

))

From (9.59) and (9.60), following inequality is observed:∣∣∣(gF , ,c,q,ra+, , , ( f ∗ g)

)(x,w; p)

∣∣∣≤ Kax (E ,c,q,r

, , ,g;) (9.61)

×((

m| f ′ ( x

m

) |e( x

m ) g(x)− | f ′(a)|ea g(a)

)

− (s+1)(x−a)s

(m| f ′ ( x

m

) |e( x

m )sJx−g(a)− | f ′(a)|

easJa+g(x)

)).

Now using exponentially (s,m)-convexity of | f ′|, we have

| f ′(t)| ≤(

t− xb− x

)s | f ′(b)|eb

+m

(b− tb− x

)s | f ′ ( xm

) |e( x

m ) . (9.62)


On the same pattern as we did for (9.6) and (9.56), one can obtain following inequalityfrom (9.11) and (9.62):∣∣∣(gF

,, ,k,c, ,l,b− ( f ∗ g)

)(x,w; p)

∣∣∣≤ Kxb(E

,c,q,r , , ,g;) (9.63)

×((

| f ′(b)|eb g(b)−m

| f ′( xm

) |e( x

m ) g(x)

)

− (s+1)(x−a)s

(| f ′(b)|eb

sJb−g(x)−m| f ′ ( x

m

) |e( x

m )sJx+g(b)

)).

By adding (9.61) and (9.63), inequality (9.55) can be achieved. �

Remark 9.3

(i) If we take = 0 in (9.55), then we get [96, Theorem 2].

(ii) If we take (s,m) = (1,1) in (9.55), then we get [155, Theorem 3].

(iii) If we take (s,m) = (1,1) and = 0 in (9.55), then we get [98, Theorem 3].

Theorem 9.7 Under the assumptions of Theorem 1, the following inequality holds form-convex functions:(

gF , ,c,q,ra+, , , f

)(x,w; p)+

(gF

, ,c,q,rb−, , , f

)(x,w; p) (9.64)

≤ Kax (E ,c,q,r

, , ,g;)(g(x)−g(a))(mf

( xm

)+ f (a)

)+Kx

b(E ,c,q,r , , ,g;)(g(b)−g(x))

(mf

( xm

)+ f (b)

).

Proof. If we put s = 1 in (9.38), we have∫ x

aKt

x(E ,c,q,r , , ,g;) f (g(t))d(g(t)) ≤ f (a)Ka

x (E ,c,q,r , , ,g;) (9.65)

×∫ x

a

(x− tx−a

)d(g(t))+mf

( xm

)Ka

x (E ,c,q,r , , ,g;)

∫ x

a

(t−ax−a

)d(g(t)).

Further simplification of (9.65), the following inequality holds:(gF

, ,c,q,ra+, , , f

)(x,w; p) ≤ Ka

x (E ,c,q,r , , ,g;)(g(x)−g(a))

(mf

( xm

)+ f (a)

). (9.66)

Similarly from (9.41), the following inequality holds:(gF

,, ,k,c, ,l,b− f

)(x,w; p) ≤ Kx

b(E, ,k,c, ,l ,g;)(g(b)−g(x))

(mf

( xm

)+ f (b)

). (9.67)

From (9.66) and (9.67), (9.64) can be obtained. �


Corollary 9.19 If we take m = 1 in Theorem 4, then the following inequality holds forconvex functions: (

gF , ,c,q,ra+, , , f

)(x,w; p)+

(gF

, ,c,q,rb−, , , f

)(x,w; p) (9.68)

≤ Kax (E ,c,q,r

, , ,g;)(g(x)−g(a))( f (x)+ f (a))

+Kxb(E

,c,q,r , , ,g;)(g(b)−g(x))( f (x)+ f (b)) .

Theorem 9.8 With assumptions of Theorem 4, if f ∈ L[a,b], then unified integral oper-ators for m-convex functions are bounded and continuous.

Proof. From (9.66) we have∣∣∣(gF , ,c,q,ra+, , , f

)(x,w; p)

∣∣∣≤ Kab (E ,c,q,r

, , ,g;)(g(b)−g(a))(m+1)‖ f‖,

which further gives ∣∣∣(gF , ,c,q,ra+, , , f

)(x,w; p)

∣∣∣≤ K‖ f‖,

where K = (g(b)−g(a))(m+1)Kab(E

,c,q,r , , ,g;).

Similarly, from (9.67) the following inequality holds:∣∣∣(gF , ,c,q,rb−, , , f

)(x,w; p)

∣∣∣≤ K‖ f‖.

�

Corollary 9.20 If we take m = 1 in Theorem 5, then unified integral operators for convexfunctions are bounded and continuous and following inequalities hold:∣∣∣(gF

, ,c,q,ra+, , , f

)(x,w; p)

∣∣∣≤ K‖ f‖,

where K = 2(g(b)−g(a))Kab(E

,c,q,r , , ,g;) and∣∣∣(gF , ,c,q,rb−, , , f

)(x,w; p)

∣∣∣≤ K‖ f‖.


9.3 Bounds of Unified Integral Operatorsfor Strongly (s,m)-Convex Functions

In this section the notation I(a,b,g) =:1

b−a

∫ ba g(t)dt is used frequently.

Theorem 9.9 Let f : [a,mb]→R, 0≤ a< mb be a positive integrable and strongly (s,m)-convex function, m �= 0. Then for unified integral operators, the following inequality holds:(

gF , ,c,q,ra+, , , f

)(x,w; p)+

(gF

, ,c,q,rb−, , ′, f

)(x,w; p) (9.69)

≤ Kax (E ,c,q,r

, , ,g;)(mf

( xm

)g(x)− f (a)g(a)

− (s+1)(x−a)s

(mf

( xm

)sJx−g(a)− f (a) sJa+g(x)

)

+ (x−ma)2

(x−a)(2I(a,x, Idg)− (a+ x)I(a,x,g))

)

+Kxb(E

,c,q,r , ′, ,g;)

(f (b)g(b)−mf

( xm

)g(x)

− (s+1)(b− x)s

(f (b) sJb−g(x)−mf

( xm

)sJx+g(b)

)

+ (mb− x)2

(b− x)(2I(x,b, Idg)− (x+b)I(x,b,g))

).

Proof. For strongly (s,m)-convex function the following inequalities hold for a < t < xand x < t < b receptively:

f (t) ≤(

x− tx−a

)s

f (a)+m

(t−ax−a

)s

f

(xm

)− (x− t)(t−a)(x−ma)2

m2(x−a)2 , (9.70)

f (t) ≤(

t− xb− x

)s

f (b)+m

(b− tb− x

)s

f

(xm

)− (t − x)(b− t)(mb− x)2

m2(b− x)2 . (9.71)

From (9.6) and (9.70), one can have

∫ x

aKt

x(E ,c,q,r , , ,g;) f (t)d(g(t)) ≤ f (a)Ka

x (E ,c,q,r , , ,g;) (9.72)

×∫ x

a

(x− tx−a

)s

d(g(t))+mf( x

m

)Ka

x (E ,c,q,r , , ,g;)

∫ x

a

(t−ax−a

)s

d(g(t))

−Kax (E ,c,q,r

, , ,g;) (x−ma)2

(x−a)2

∫ x

a(x− t)(t−a)d(g(t)),


i.e, (gF

, ,c,q,ra+, , , f

)(x,w; p) ≤ Ka

x (E ,c,q,r , , ,g;)

(mf

( xm

)g(x)− f (a)g(a) (9.73)

− (s+1)(x−a)s

(mf

( xm


)

+ (x−ma)2


).

On the other hand from (9.11) and (9.71), one can have∫ b

xKx

t (E ,c,q,r , ′, ,g;) f (t)d(g(t)) ≤ f (b)Kx

b(E ,c,q,r , ′, ,g;) (9.74)

×∫ b

x

(t− xb− x

)s

d(g(t))+mf( x

m

)Kb

x (E ,c,q,r , ′, ,g;)

∫ b

x

(b− tb− x

)s

d(g(t))

−Kbx (E ,c,q,r

, , ,g;) (mb− x)2

m2(b− x)2

∫ b

x(t− x)(b− t)d(g(t)),

i.e, (gF

, ,c,q,rb−, , ′, f

)(x,w; p) ≤ Kx

b(E ,c,q,r , ′, ,g;)

(f (b)g(b)−mf

( xm

)g(x) (9.75)

− (s+1)(b− x)s


( xm

)sJx+g(b)

)

+ (mb− x)2


).


Corollary 9.21 Setting p = w = 0 in (9.69) we can obtain the following inequality in-volving fractional integral operators defined in [49]:(

F ,a+ f

)(x; p)+

(F ′,b− f

)(x; p) (9.76)

≤ Kg(a,x;)(mf

( xm

)g(x)− f (a)g(a)

− (s+1)(x−a)s

(mf

( xm


)

+ (x−ma)2


)

+Kg(x,b;)(

f (b)g(b)−mf( x

m

)g(x)

− (s+1)(b− x)s


( xm

)sJx+g(b)

)

+ (mb− x)2


).


Remark 9.4

(i) If we consider = 0 in (9.69), then [69, Theorem 3.1] can be obtained, for > 0we get its refinement.

(ii) If we consider (t) = t and g(x) = x in (9.69), then [55, Theorem 1] can be ob-tained.

(iii) If we consider s = m = 1 in the result of (ii), then [55, Corollary 1] can be obtained.

(iv) If we consider = ′ in the result of (ii), then [55, Corollary 3] can be obtained.

(v) If we consider f ∈ L[a,b] in the result of (ii), then [55, Corollary 5] can be obtained.

(vi) If we consider = ′ in the result of (v), then [55, Corollary 7] can be obtained.

(vii) If we consider s = 1 in the result of (ii), then [55, Corollary 5] can be obtained.

(viii) If we consider (s,m)= (1,1) in (9.69), then [82, Theorem 2] is obtained.

(ix) If we consider = ′, = 0 and (s,m)= (1,1) in (9.69), then [98, Theorem 8] isobtained.

(x) If we consider = 0 and p = w = 0 in (9.69), then [96, Theorem 1] is obtained.

(xi) If we consider = 0, (t) = ()t , p = w = 0 and (s,m)= (1,1) in (9.69), then[59, Theorem 1] is obtained.

(xii) If we consider = ′ in the result of (xi), then [59, Corollary 1] is obtained.

(xiii) If we consider = 0, (t) = t , g(x) = x and m = 1 in (9.69), then [35, Theorem2.1] is obtained.

(xvi) If we consider = ′ in the result of (xiii), then [35, Corollary 2.1] is obtained.

(xv) If we consider = 0, (t) = ()tk

kk() , (s,m)= (1,1), g(x) = x and p = w = 0 in (9.69),then [48, Theorem 1] can be obtained.

(xvi) If we consider = ′ in the result of (xv), then [48, Corollary 1] can be obtained.

(xvii) If we consider = 0, (t) = ()t , g(x) = x and p = w = 0 and (s,m)= (1,1) in(9.69), then [47, Theorem 1] is obtained.

(xviii) If we consider = ′ in the result of (xvii), then [47, Corollary 1] can be obtained.

(xviii) If we consider = ′ = 1 and x = a or x = b in the result of (xvii), then [47,Corollary 2] can be obtained.

(xix) If we consider = ′ = 1 and x =a+b

2in the result of (xvii), then [47, Corollary

3] can be obtained.

The following lemma is very helpful in the proof of upcoming theorem, see [55].

Lemma 9.3 Let f : [a,mb] → R be strongly strongly (s,m)-convex function, 0 ≤ a < mb.If f is f ( a+mb−x

m ) = f (x), m �= 0, then the following inequality holds:

f

(a+mb

2

)≤ (1+m) f (x)

2s − 4m

(a+mb− x−mx)2. (9.77)

By using above lemma, the following inequality for strongly (s,m)-convex functions isobtained.


Theorem 9.10 Under the assumptions of Theorem 9.9, in addition if f (x)= f

(a+mb− x

m

),

then the following inequality holds:

f

(a+mb

2

)2s

(1+m)

((gF

, ,c,q,ra+, , ′,1

)(b,w; p) (9.78)

+4m

(gF

, ,c,q,rb−, , , (a+mb− x−mx)2

)(a,w; p)+

(gF

, ,c,q,rb−, , , 1

)(a,w; p)

+4m

(gF

, ,c,q,ra+, , ′,(a+mb− x−mx)2

)(b,w; p)

)

≤(

gF , ,c,q,ra+, , ′, f

)(b,w; p)+

(gF

, ,c,q,rb−, , , f

)(a,w; p)

≤(Ka

b (E ,c,q,r , , ,g;)+Ka

b (E ,c,q,r , ′, ,g;)

)(f (b)g(b)−mf

(am

)g(a)

− (s+1)(b−a)s

(f (b) sJb−g(a)−mf

(am

)sJa+g(b)

)

+ (mb−a)2

(b−a)(2I(a,b, Idg)− (a+b)I(a,b,g))

).

Proof. For strongly (s,m)-convex function satisfies the following inequalities hold fora < x < b:

f (x) ≤(

x−ab−a

)s

f (b)+m

(b− xb−a

)s

f

(am

)− (b− x)(x−a)(b−ma)2

m2(b−a)2 . (9.79)

From (9.16) and (9.79), one can have∫ b

aKa

x (E ,c,q,r , , ,g;) f (x)d(g(x))

≤ mf( a

m

)Ka

b (E ,c,q,r , , ,g;)

∫ b

a

(b− xb−a

)s

d(g(x))

+ f (b)Kab (E ,c,q,r

, , ,g;)∫ b

a

(x−ab−a

)s

d(g(x))

−Kba (E ,c,q,r

, , ,g;) (b−ma)2

m2(b−a)2

∫ b

a(x−a)(b− x)d(g(x)).

Further, the aforementioned inequality takes the form which involves Riemann-Liouvillefractional integrals in the right hand side, thus we have upper bound of the unified leftsided integral operator as follows:(

gF , ,c,q,rb−, , , f

)(a,w; p) ≤ Ka

b (E ,c,q,r , , ,g;)

(f (b)g(b)−mf

(am

)g(a) (9.80)

− (s+1)(b−a)s


(am

)sJa+g(b)

)

+ (mb−a)2


).


On the other hand

Kxb(E

,c,q,r , ′, ,g;)g′(x) ≤ Ka

b (E ,c,q,r , ′, ,g;)g′(x) a < x < b, (9.81)

from (9.81) and (9.79), the following inequality holds which involves Riemann-Liouvillefractional integrals on the right hand side, gives estimate of the unified right sided integraloperator:(

gF , ,c,q,ra+, , ′, f

)(b,w; p) ≤ Ka

b (E ,c,q,r , ′, ,g;)

(f (b)g(b)−mf

(am

)g(a) (9.82)

− (s+1)(b−a)s


(am

)sJa+g(b)

)

+ (mb−a)2


).


, ,c,q,rb−, , , f

)(a,w; p)+

(gF

, ,c,q,ra+, , ′, f

)(b,w; p) (9.83)

≤(Ka

b (E ,c,q,r , , ,g;)+Ka

b (E ,c,q,r , ′, ,g;)

)(f (b)g(b)−mf

(am

)g(a)

− (s+1)(b−a)s


(am

)sJa+g(b)

)

+ (mb−a)2


).


, , ,g;)g′(x), and integrating over [a,b] wehave

f

(a+b

2

)∫ b

aKa

x (E ,c,q,r , , ,g;)d(g(x))

≤(

12s

)(1+m)

∫ b

aKa

b (E ,c,q,r , , ,g;) f (x)d(g(x))

− 4m

∫ b

aKa

x (E ,c,q,r , , ,g;)(a+mb− x−mx)2d(g(x)).

From which the following inequality is obtained:

f

(a+mb

2

)2s

(1+m)

(gF

, ,c,q,rb−, , , 1

)(a,w; p) ≤

(gF

, ,c,q,rb−, , , f

)(a,w; p) (9.84)

− 4m

(gF

, ,c,q,rb−, , , (a+mb− x−mx)2

)(a,w; p).


,c,q,r , ′, ,g;)g′(x), and integrating over

[a,b] we have

f

(a+mb

2

)2s

(1+m)

(gF

, ,c,q,ra+, , ′,1

)(b,w; p) ≤

(gF

, ,c,q,ra+, , ′, f

)(b,w; p) (9.85)

− 4m

(gF

, ,c,q,ra+, , ′, (a+mb− x−mx)2

)(b,w; p).


By adding (9.84) and (9.85) the following inequality is obtained:

f

(a+mb

2

)2s

(1+m)

((gF

, ,c,q,ra+, , ′,1

)(b,w; p) (9.86)

+4m

(gF

, ,c,q,rb−, , , (a+mb− x−mx)2

)(a,w; p)+

(gF

, ,c,q,rb−, , , 1

)(a,w; p)

+4m

(gF

, ,c,q,ra+, , ′,(a+mb− x−mx)2

)(b,w; p)

)

≤(

gF , ,c,q,ra+, , ′, f

)(b,w; p)+

(gF

, ,c,q,rb−, , , f

)(a,w; p).

Using (9.83) and (9.86), inequality (9.78) can be obtained. �

Corollary 9.22 Setting p = w = 0 in (9.78) we can obtain the following inequality in-volving fractional integral operators defined in [49]:

f

(a+mb

2

)2s

(1+m)

((F ,a+1

)(b; p) (9.87)

+4m

(F ,b−(a+mb− x−mx)2

)(a; p)+

(F ,b−1

)(a; p)

+4m

(F ′,a+(a+mb− x−mx)2

)(b; p)

)

≤(F ,a+ f

)(b; p)+

(F ,b− f

)(a; p)

≤ (Kg(a,b;)+Kg(a,b;))(

f (b)g(b)−mf

(am

)g(a)

− (s+1)(b−a)s


(am

)sJa+g(b)

)

+ (mb−a)2


).

Remark 9.5

(i) If we consider (t) = t and g(x) = x in (9.78), then [55, Theorem 7] can be ob-tained.

(ii) If we consider = 0 in the result of (i), then [55, Theorem 8] can be obtained.

(iii) If we consider (s,m)= (1,1) in (9.78), then [82, Theorem 3] is obtained.

(iv) If we consider = 0 and (s,m)= (1,1) in (9.78), then [98, Theorem 22] is obtained.

(v) If we consider = 0, (t) = ()t+1, p = w = 0 and (s,m)= (1,1) in (9.78), then[59, Theorem 3] is obtained.

(vi) If we consider = ′ in the result of (v), then [59, Corollary 3] is obtained.

(vii) If we consider = 0, (t) = t+1, g(x) = x and m = 1 in (9.78), then [35, Theorem2.4] is obtained.


(viii) If we consider = ′ in the result of (vii), then [35, Corollary 2.6] is obtained.

(ix) If we consider = 0, (t) = ()tk +1, (s,m)= (1,1), g(x) = x and p = w = 0 in

(9.78), then [48, Theorem 3] can be obtained.

(x) If we consider = ′ in the result of (ix), then [48, Corollary 6] can be obtained.

(xi) If we consider = 0, (t) = ()t+1, p = w = 0, (s,m) = 1 and g(x) = x in (9.78),then [47, Theorem 3] can be obtained.

(xii) If we consider = ′ in the result of (xi), then [47, Corrolary 6] can be obtained.

Theorem 9.11 Let f : [a,mb] → R, 0 ≤ a < mb be differential function such that | f ′| isstrongly (s,m)-convex function, m �= 0. Then for unified integral operators, the followinginequality holds:∣∣∣(gF

, ,c,q,ra+, , , f ∗ g

)(x,w; p)+

(gF

, ,c,q,rb−, , ′, f ∗ g

)(x,w; p)

∣∣∣ (9.88)

≤ Kax (E ,c,q,r

, , ,g;)((

m∣∣∣ f ′( x

m

)∣∣∣g(x)−| f ′(a)|g(a)

− (s+1)(x−a)s

(m∣∣∣ f ′( x

m

)∣∣∣ sJx−g(a)−| f ′(a)| sJa+g(x))

+ (x−am)2


)

+Kxb(E

,c,q,r , ′, ,g;)

((| f ′(b)|g(b)−m

∣∣∣ f ′( xm

)∣∣∣g(x))

− (s+1)(b− x)s

(| f ′(b)| sJb−g(x)−m

∣∣∣ f ′( xm

)∣∣∣ sJx+g(b))

+ (mb− x)2


).

Proof. For strongly (s,m)-convex function | f ′|, the following inequalities hold for a < t <x and x < t < b receptively:

| f ′(t)| ≤(

x− tx−a

)s

| f ′(a)|+m

(t−ax−a

)s ∣∣∣∣ f ′(

xm

)∣∣∣∣− (x− t)(t−a)(x−ma)2

m2(x−a)2 , (9.89)

| f ′(t)| ≤(

t− xb− x

)s

| f ′(b)|+m

(b− tb− x

)s ∣∣∣∣ f ′(

xm

)∣∣∣∣− (t− x)(b− t)(mb− x)2

m2(b− x)2 . (9.90)

From (9.6) and (9.89), the following inequality is obtained:∣∣∣(gF , ,c,q,ra+, , , ( f ∗ g)

)(x,w; p)

∣∣∣≤ Kax (E ,c,q,r

, , ,g;)((

m∣∣∣ f ′( x

m

)∣∣∣g(x)−| f ′(a)|g(a)

(9.91)

− (s+1)(x−a)s

(m∣∣∣ f ′( x

m

)∣∣∣ sJx−g(a)−| f ′(a)| sJa+g(x))

+ (x−am)2


).


Similarly, from (9.11) and (9.90), the following inequality is obtained:∣∣∣(gF , ,c,q,rb−, , ′,( f ∗ g)

)(x,w; p)

∣∣∣≤ Kxb(E

,c,q,r , ′, ,g;)

((| f ′(b)|g(b)−m

∣∣∣ f ′( xm

)∣∣∣g(x))

(9.92)

− (s+1)(b− x)s

(| f ′(b)| sJb−g(x)−m

∣∣∣ f ′( xm

)∣∣∣ sJx+g(b))

+ (mb− x)2


).


Corollary 9.23 Setting p = w = 0 in (9.88) we can obtain the following inequality in-volving fractional integral operators defined in [49]:∣∣∣(F

,a+ f ∗ g)

(x, p)+(F ′,b− f ∗ g

)(x, p)

∣∣∣ (9.93)

≤ Kg(a,x;)((

m∣∣∣ f ′( x

m

)∣∣∣g(x)−| f ′(a)|g(a)

− (s+1)(x−a)s

(m∣∣∣ f ′( x

m

)∣∣∣ sJx−g(a)−| f ′(a)| sJa+g(x))

+ (x−am)2


)

+Kb(x,b;)((

| f ′(b)|g(b)−m∣∣∣ f ′( x

m

)∣∣∣g(x))

− (s+1)(b− x)s

(| f ′(b)| sJb−g(x)−m

∣∣∣ f ′( xm

)∣∣∣ sJx+g(b))

+ (mb− x)2


).

Remark 9.6

(i) If we consider = 0 in (9.88), then [69, Theorem 3.4] can be obtained.

(ii) If we consider (t) = t and g(x) = x in (9.88), then [55, Theorem 6] can be ob-tained.

(iii) If we consider s = m = 1 in the result of (ii), then [55, Corollary 13] can be obtained.

(iv) If we consider = ′ in the result of (ii), then [55, Corollary 11] can be obtained.

(v) If we consider (s,m)= (1,1) in (9.88), then [82, Theorem 3] is obtained.

(vi) If we consider = 0 and (s,m)= (1,1) in (9.88), then [98, Theorem 25] is obtained.

(vii) If we consider = 0 and p = w = 0 in (9.88), then [96, Theorem 2] is obtained.

(viii) If we consider = 0, (t) = ()t+1, p = w = 0 and (s,m)= (1,1) in (9.88), then[59, Theorem 2] is obtained.

(ix) If we consider = ′ in the result of (viii), then [59, Corollary 2] is obtained.

9.4 BOUNDS OF OPERATORS FOR (,m)-CONVEX FUNCTIONS 253

(x) If we consider = 0, (t) = t , g(x) = x and m = 1 in (9.88), then [35, Theorem2.3] is obtained.

(xi) If we consider = ′ in the result of (x), then [35, Corollary 2.5] is obtained.

(xii) If we consider = 0, (t) = ()tk +1, (s,m)= (1,1), g(x) = x and p = w = 0 in

(9.88), then [48, Theorem 2] can be obtained.

(xiii) If we consider = ′ in the result of (xii), then [48, Corollary 4] can be obtained.

(xiv) If we consider = ′ = k = 1 and x =a+b

2, in the result of (xii), then [48, Corollary

5] can be obtained.

(xv) If we consider = 0, (t) = ()t+1, g(x) = x and p = w = 0 and (s,m)= (1,1) in(9.88), then [47, Theorem 2] is obtained.

(xvi) If we consider = ′ in the result of (xv), then [47, Corollary 5] can be obtained.

9.4 Bounds of Unified Integral Operatorsfor (,m)-Convex Functions

In this section bounds of unified integral operators of (,m)-convex functions are estab-lished.

Theorem 9.12 Let f : [a,b] → R be a positive integrable (,m)-convex function withm ∈ (0,1]. Let g : [a,b] → R be differentiable and strictly increasing function, also let

xbe an increasing function on [a,b]. If , l,,c ∈ C, (),(l) > 0, (c) > () > 0,p, , ≥ 0 and 0 < k ≤ + , then for x ∈ (a,b) we have(

gF , ,c,q,ra+, , , f

)(x,w; p) ≤ Ka

x (E ,c,q,r , , ,g;)

((mf

( xm

)g(x)− f (a)g(a)

)(9.94)

−(+1)(x−a)

(mf

( xm

)− f (a)

)Ja+g(x)

),

(gF

, ,c,q,rb−, , , f

)(x,w; p) ≤ Kx

b(E ,c,q,r , , ,g;)

((f (b)g(b)−mf

( xm

)g(x)

)(9.95)

( +1)(b− x)

(f (b)−mf

( xm

))Jb−g(x)

)

and hence (gF

, ,c,q,ra+, , , f

)(x,w; p)+

(gF

, ,c,q,rb−, , , f

)(x,w; p) (9.96)

≤ Kax (E ,c,q,r

, , ,g;)((

mf( x

m

)g(x)− f (a)g(a)

)−(+1)

(x−a)(mf

( xm

)− f (a)

)Ja+g(x)

)


+Kxb(E

,c,q,r , , ,g;)

((f (b)g(b)−mf

( xm

)g(x)

)−( +1)

(b− x)(

f (b)−mf( x

m

))Jb−g(x)

).

Proof. Using definition of (,m)-convexity for f the following inequality is valid:

f (t) ≤(

x− tx−a

)f (a)+m

(1−

(x− tx−a

))f

(xm

). (9.97)

Multiplying (9.6) with (9.97) and integrating over [a,x], one can obtain

∫ x

aKt

x(E ,c,q,r , , ,g;) f (t)d(g(t)) ≤ f (a)Ka

x (E ,c,q,r , , ,g;)

∫ x

a

(x− tx−a

)d(g(t))

+mf( x

m

)Ka

x (E ,c,q,r , , ,g;)

∫ x

a

(1−

(x− tx−a

))d(g(t)).


, ,c,q,ra+, , , f

)(x,w; p) ≤ Ka

x (E ,c,q,r , , ,g;)

((x−a)

(mf

( xm

)g(x)− f (a)g(a)

)(9.98)

( +1)(x−a)

(mf

( xm

)− f (a)

)Ja+g(x)

).

Using (,m)-convexity of f , we have

f (t) ≤(

t− xb− x

)f (b)+m

(1−

(t− xb− x

))f

(xm

). (9.99)

Adopting the same procedure as we did for (9.6) and (9.97), the following inequality from(9.11) and (9.99) can be obtained:(

gF , ,c,q,rb−, , , f

)(x,w; p) ≤ Kx

b(E ,c,q,r , , ,g;)

((f (b)g(b)−mf

( xm

)g(x)

)(9.100)

−( +1)(b− x)

(f (b)−mf

( xm

))Jb−g(x)

).


Remark 9.7

(i) If we consider (,m) =(1,1) in (9.96), [98, Theorem 1] is obtained.

(ii) If we consider (t) = ()tk

kk() for left hand integral and (t) = ()tk

kk() for right handintegral and p = w = 0 in (9.96), then [88, Theorem 1] can be obtained.

(iii) If we consider = in the result of (ii), then [88, corollary 1] can be obtained.

(iv) If we consider (t) = ()t , p = w = 0 and (,m) =(1,1) in (9.96), [59, Theorem1] is obtained.


(v) If we consider = in the result of (iv), [59, Corollary 1] is obtained.

(vi) If we consider (t) = ()tk

kk() for left hand integral and (t) = ()tk

kk() for right hand

integral, (,m) =(1,1), g(x) = x and p = w = 0, then [48, Theorem 1] can be ob-tained.

(vii) If we consider = in the result of (vi), then [48, Corollary 1] can be obtained.

(viii) If we consider (t) = ()t for left hand integral and (t) = ()t for right handintegral, g(x) = x and p = w = 0 and (,m) =(1,1) in (9.96), then [47, Theorem 1]is obtained.

(ix) By setting = in the result of (viii), [47, Corollary 1] can be obtained.

(x) By setting = = 1 and x = a or x = b in the result of (ix), [47, Corollary 2] canbe obtained.

(xi) By setting = = 1 and x =a+b

2in the result of (ix), [47, Corollary 3] can be

obtained.

To prove the the next result we need the following lemma [88].

Lemma 9.4 Let f : [0,] → R, be an (,m)-convex function with m ∈ (0,1]. If f (x) =f ( a+b−x

m ),0 < a < b, then the following inequality holds:

f

(a+b

2

)≤ 1

2(1+m(2 −1)) f (x), (9.101)

for all x ∈ [a,b] and m ∈ (0,1].

Theorem 9.13 With the assumptions of Theorem 9.12 in addition if f (x)= f

(a+b− x

m

),

then we have

2 f

(a+b

2

)(1+m(2−1))

((gF

, ,c,q,rb−, , , 1

)(a,w; p)+

(gF

, ,c,q,ra+, , , 1

)(b,w; p)

)(9.102)

≤(

gF , ,c,q,rb−, , , f

)(a,w; p)+

(gF

, ,c,q,ra+, , , f

)(b,w; p)

≤ 2Kab (E ,c,q,r

, , ,g;)((

f (b)g(b)−mf( a

m

)g(a)

)−(+1)

(b−a)(

f (b)−mf( a

m

))Jb−g(a)

).

Proof. Using (,m)-convexity of f for x ∈ (a,b), we have

f (x) ≤(

x−ab−a

)f (b)+m

(1−

(x−ab−a

))f

(am

). (9.103)


Multiplying (9.16) and (9.103) and integrating the resulting inequality over [a,b], one canobtain ∫ b

aKa

x (E ,c,q,r , , ,g;) f (x)d(g(x))

≤ mf( a

m

)Ka

b (E ,c,q,r , , ,g;)

∫ b

a

(1−

(x−ab−a

))d(g(x))

+ f (b)(g(b)−g(a))

g(b)−g(a)E ,c,q,r , , (w(g(b)−g(a)) ; p)

∫ b

a

(x−ab−a

)d(g(x)).

By using Definition 2.3 and integrating by parts, the following inequality is obtained:

(gF

, ,c,q,ra+, , , f

)(b,w; p) ≤ Ka

b (E ,c,q,r , , ,g;)

((f (b)g(b)−mf

(am

)g(a)

)(9.104)

− ( +1)(b−a)

(f (b)−mf

(am

))Jb−g(a)

).

Adopting the same pattern of simplification as we did for (9.16) and (9.103), the followinginequality can be observed from (9.103) and (9.19)

(gF

, ,c,q,rb−, , , f

)(a; p) ≤ Ka

b (E ,c,q,r , , ,g;)

((f (b)g(b)−mf

(am

)g(a)

)(9.105)

− ( +1)(b−a)

(f (b)−mf

(am

))Jb−g(a)

).


, ,c,q,rb−, , , f

)(a,w; p)+

(gF

, ,c,q,ra+, , , f

)(b,w; p) (9.106)

≤ Kab (E ,c,q,r

, , ,g;)((

f (b)g(b)−mf( a

m

)g(a)

)−(+1)

(b−a)(

f (b)−mf( a

m

))Jb−g(a)

).


, , ,g;)d(g(x)), and integrating over [a,b]we have

f

(a+b

2

)∫ b

aKa

x (E ,c,q,r , , ,g;)d(g(x)) ≤

(12

)(1+m(2 −1))

×∫ b

aKa

x (E ,c,q,r , , ,g;) f (x)d(g(x)).

From Definition 2.3, the following inequality is obtained:

f

(a+b

2

)2

(1+m(2 −1))

(gF

, ,c,q,rb−, , , 1

)(a; p) ≤

(gF

, ,c,q,rb−, , , f

)(a,w; p). (9.107)



,c,q,r , , ,g;)d(g(x)), and integrating

over [a,b] we have

f

(a+b

2

)2

(1+m(2−1))

(gF

, ,c,q,ra+, , , 1

)(b; p) ≤

(gF

, ,c,q,ra+, , , f

)(b,w; p). (9.108)

By adding (9.107) and (9.108) following inequality is obtained:

f

(a+b

2

)2

(1+m(2−1))

((gF

, ,c,q,rb−, , , 1

)(a,w; p)+

(gF

, ,c,q,ra+, , , 1

)(b,w; p)

)(9.109)

≤(

gF , ,c,q,rb−, , , f

)(a,w; p)+

(gF

, ,c,q,ra+, , , f

)(b,w; p).

Using (9.106) and (9.109), inequality (9.102) can be achieved. �

Remark 9.8

(i) If we consider (,m) =(1,1) in (9.102), [98, Theorem 2] is obtained.

(ii) If we consider (t) = ()tk +1 for left hand integral and (t) = ()t

k +1 and

p = w = 0 in (9.102), then [88, Theorem 3] can be obtained.

(iii) If we consider = in the result of (ii), then [88, corollary 3] can be obtained.

(iv) If we consider (t) = ()t+1 for left hand integral and (t) = ()t+1for righthand integral in (9.102), p = w = 0 and (,m) =(1,1) in (9.102), [59, Theorem 3] isobtained.

(v) If we consider = in the result of (iv), [59, Corollary 3] is obtained.

(vi) If we consider (t) = ()tk +1 for left hand integral and (t) = ()t

k +1for right

hand integral, (,m) =(1,1), g(x) = x and p = w = 0 in (9.102), then [48, Theorem3] can be obtained.


(viii) By setting (t) = ()t+1 for left hand integral and (t) = ()t+1 for righthand integral p = w = 0, (,m) = 1 and g(t) = t in (9.102), [47, Theorem 3] can beobtained.

(ix) By setting = in the result of (viii), [47, Corrolary 6] can be obtained.

Theorem 9.14 Let f : [a,b] → R be a differentiable function. If | f ′| is (,m)-convexwith m ∈ (0,1] and g : [a,b]→ R be differentiable and strictly increasing function, also letx be an increasing function on [a,b]. If , l,,c ∈ C, (),(l) > 0, (c) > () > 0,p, , ≥ 0 and 0 < k ≤ + , then for x ∈ (a,b) we have∣∣∣(gF

, ,c,q,ra+, , , f

)(x,w; p)+

(gF

, ,c,q,rb−, , , f

)∣∣∣(x,w; p) (9.110)

≤ Kax (E ,c,q,r

, , ,g;)((

m∣∣∣ f ′( x

m

)∣∣∣g(x)−| f ′(a)|g(a))

−( +1)(x−a)

(m∣∣∣ f ′( x

m

)∣∣∣−| f ′(a)|)

Ja+g(x))


+Kxb(E

,c,q,r , , ,g;)

((| f ′(b)|g(b)−m

∣∣∣ f ′( xm

)∣∣∣g(x))

− ( +1)(b− x)

(| f ′(b)|−m

∣∣∣ f ′( xm

)∣∣∣) Jb−g(x))

,

Proof. Let x ∈ (a,b) and t ∈ [a,x]. Then using (,m)-convexity of | f ′| we have

| f ′(t)| ≤(

x− tx−a

)| f ′(a)|+m

(1−

(x− tx−a

))∣∣∣∣ f ′(

xm

)∣∣∣∣ . (9.111)

The inequality (9.111) can be written as follows:

−((

x− tx−a

)| f ′(a)|+m

(1−

(x− tx−a

))∣∣∣∣ f ′(

xm

)∣∣∣∣)≤ f ′(t) (9.112)

≤(

x− tx−a

)| f ′(a)|+m

(1−

(x− tx−a

))∣∣∣∣ f ′(

xm

)∣∣∣∣ .Let we consider the second inequality of (9.112)

f ′(t) ≤(

x− tx−a

)| f ′(a)|+m

(1−

(x− tx−a

))∣∣∣∣ f ′(

xm

)∣∣∣∣ . (9.113)

Multiplying (9.6) and (9.113) and integrating over [a,x], we can obtain:

∫ x

aKt

x(E ,c,q,r , , ,g;)d(g(t)) ≤ | f (a)|Ka

x (E ,c,q,r , , ,g;)

∫ x

a

(x− tx−a

)d(g(t))

+m∣∣∣ f ( x

m

)∣∣∣Kax (E ,c,q,r

, , ,g;)∫ x

a

(1−

(x− tx−a

))d(g(t)).


, ,c,q,ra+, , , f

)(x,w; p) ≤ Ka

x (E ,c,q,r , , ,g;) (9.114)

×((

m∣∣∣ f ′( x

m

)∣∣∣g(x)−| f ′(a)|g(a))

− ( +1)(x−a)

(m∣∣∣ f ′( x

m

)∣∣∣−| f ′(a)|)

Ja+g(x))

.

If we consider the left hand side from the inequality (9.112) and adopt the same pattern asdid for the right hand side inequality, then(

gF ,, ,k,c, ,l,a+ ( f ∗ g)

)(x,w; p) ≥−Ka

x (E ,c,q,r , , ,g;) (9.115)

×((

m∣∣∣ f ′( x

m

)∣∣∣g(x)−| f ′(a)|g(a))

− ( +1)(x−a)

(m∣∣∣ f ′( x

m

)∣∣∣−| f ′(a)|)

Ja+g(x))

.


From (9.114) and (9.115), following inequality is observed:∣∣∣(gF ,, ,k,c, ,l,a+ ( f ∗ g)

)(x,w; p)

∣∣∣≤ Kax (E ,c,q,r

, , ,g;) (9.116)

×((

m∣∣∣ f ′( x

m

)∣∣∣g(x)−| f ′(a)|g(a))

− ( +1)(x−a)

(m∣∣∣ f ′( x

m

)∣∣∣−| f ′(a)|)

Ja+g(x))

.

Now using (,m)-convexity of | f ′| on (x,b] for x ∈ (a,b) we have

| f ′(t)| ≤(

t − xb− x

)| f ′(b)|+m

(1−

(t− xb− x

))∣∣∣∣ f ′(

xm

)∣∣∣∣ . (9.117)

On the same procedure as we did for (9.6) and (9.111), one can obtain following inequalityfrom (9.11) and (9.117):∣∣∣(gF

,, ,k,c, ,l,b− ( f ∗ g)

)(x,w; p)

∣∣∣≤ Kxb(E

,c,q,r , , ,g;) (9.118)

×((

| f ′(b)|g(b)−m∣∣∣ f ′( x

m

)∣∣∣g(x))

− ( +1)(b− x)

(| f ′(b)|−m

∣∣∣ f ′( xm

)∣∣∣) Jb−g(x))

.


Remark 9.9(i) If we consider (,m) =(1,1) in (9.110), then [98, Theorem 3] is obtained.

(ii) If we consider (t) = ()tk +1 for left hand integral and (t) = ()t

k +1 for right

hand integral and p = w = 0 in (9.110), then [88, Theorem 2] can be obtained.

(iii) If we consider = in the result of (ii), then [88, Corollary 2] can be obtained.

(iv) If we consider (t) = ()t+1 for left hand integral and (t) = ()t+1 for righthand integral, p = w = 0 and (,m) =(1,1) in (9.110), then [59, Theorem 2] isobtained.

(v) If we consider = in the result of (iv), then [59, Corollary 2] is obtained.

(vi) If we consider (t) = ()tk +1 for left hand integral and (t) = ()t

k +1 for right

hand integral, (,m) =(1,1), g(x) = x and p = w = 0 in (9.110), then [48, Theorem2] can be obtained.


(viii) If we consider = = k = 1 and x =a+b

2, in the result of (vii)], then [48, Corollary

5] can be obtained.

(ix) If we consider (t) = ()t+1 for left hand integral and (t) = ()t+1 for righthand integral, g(x) = x and p = w = 0 and (,m) =(1,1) in (9.110), then [47, Theo-rem 2] is obtained.

(x) By setting = in the result of (ix), then [47, Corollary 5] can be obtained.

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Index

(,m)-convex function, 11(h−m)-convex function, 10(s,m)-convex function, 11m-convex function, 10n-convex function, 7p-convex function, 12p-symmetric function, 12s-convex function, 11

absolutely continuous function, 2

beta function, 4

Chebyshev inequality, 37concave function, 5continuous function, 1convex function, 5

divided difference, 7

exponentially (s,m)-convex function, 11exponentially m-convex function, 10exponentially s-convex function, 11

Fejer-Hadmard inequality, 82Fubini theorem, 2function

— (,m)-convex, 11— (h−m)-convex, 10— (s,m)-convex, 11— m-convex, 10— p-convex, 12— p-symmetric, 12— s-convex, 11— exponentially (s,m)-convex, 11— exponentially m-convex, 10— exponentially s-convex, 11— harmonically (,h−m)-convex, 9— harmonically (h−m)-convex, 9

— harmonically convex, 8— harmonically symmetric, 9— quasi-convex, 8— relative convex, 8— n-convex, 7— absolutely continuous, 2— beta, 4— concave, 5— continuous, 1— convex, 5— gamma, 3— log-convex, 7— Mittag-Leffler, 17— strictly concave, 5— strictly convex, 5

gamma function, 3Gruss inequality, 37

Hadamard inequality, 81harmonically (,h−m)-convex function,

9harmonically (h−m)-convex function, 9harmonically convex function, 8harmonically symmetric function, 9

inequality— Chebyshev, 37— Fejer-Hadamard, 82— Gruss, 37— Hadamard, 81— Jensen, 5— Minkowski, 49— Opial, 29— Polya-Szego, 38

log-convex function, 7

271

Minkowski inequality, 49

Opial inequality, 29

Polya-Szego inequality, 38Pochhammer symbol, 3

quasi-convex function, 8

relative convex function, 8Riemann-Liouville!fractional derivative,

13Riemann-Liouville!fractional integral, 12

space— ACn[a,b], 2— Cn[a,b], 1— L[a,b], 1— Lp, 1

Zagreb, May 2021

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Maja Andri´c, Ghulam Farid and Josip Peˇcari´ c